ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable MCQs

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15

 MCQs

Question 1

Which of the following is not a quadratic equation ?

(a) (x+2)²=2(x+3)

(b) x²+3x=(–1) (1–3x)

(c) (x+2)(x–1)=x²–2x–3

(d) x³–x²+2x+1=(x+1)3

Sol :

(a) (x + 2)² = 2(x + 3)

⇒ x² + 4x + 4 = 2x + 6

⇒ x² + 4x – 2x + 4 – 6 = 0

⇒ x² + 2x – 2

It is a quadratic equation.

(b) $x^{2}+3 x=(-1)(1-3 x) $

$\Rightarrow x^{2}+3 x=-1+3 x$

$\Rightarrow x^{2}+1=0$

It is also quadratic equation.

(c) $(x+2)(x-1)=x^{2}-2 x-3$

$x^{2}-x+2 x-2=x^{2}-2 x-3$

$x^{2}-x^{2}+x+2 x-2+3=0 \Rightarrow 3 x+1=0$

It is not a quadratic equation.

(d) $x^{3}-x^{2}+2 x+1=(x+1)^{3}$

$=x^{3}+3 x^{2}+3 x+1$

$x^{3}-x^{2}+2 x+1$

$3 x^{2}+x^{2}-2 x-1+3 x+1=0$

$\Rightarrow 4 x^{2}+x=0$

It is a quadratic equation 

Ans : (c)

Question 2

Which of the following is a quadratic equation ?

(a) (x – 2) (x + 1) = (x – 1) (x – 3)

(b) $(x+2)^{3}=2 x\left(x^{2}-1\right)$

(c) $x^{2}+3 x+1=(x-2)^{2}$

(d) $8(x-2)^{3}=(2 x-1)^{3}+3$

Sol :

(a) (x – 2) (x + 1) = (x – 1) (x – 3)
⇒ x2 + x – 2x – 2 = x2 – 3x – x + 3
⇒ 3x + x – 2x + x = 3 + 2
⇒ 3x = 5
It is not a quadratic equation.

(b) $(x+2)^{3}=2 x\left(x^{2}-1\right)$

$x^{3}+6 x^{2}+12 x+8=2 x^{3}-2 x$

$x^{3}+6 x^{2}+12 x+8-2 x^{3}+2 x=0$

$-x^{3}+6 x^{2}+14 x+8=0$

It is not a quadratic equation.

(c) $x^{2}+3 x+1=(x-2)^{2}$

$x^{2}+3 x+1=x^{2}-4 x+4$

⇒3x+1+4 x-4=0 

⇒7x-3=0

It is not a quadratic equation.

(d) $8(x-2)^{3}=(2 x-1)^{3}+3$

$8\left(x^{3}-6 x^{2}+12 x-8\right)$

$=8 x^{3}-12 x^{2}+6 x-1+3$

$8 x^{3}-48 x^{2}+96 x-64-8 x^{3}+12 x^{2}-6 x+1-3=0$

$-36 x^{2}+90 x-66=0$

It is a quadratic equation

Ans : (d)

Question 3

Which of the following equations has 2 as a root ?

(a) $x^{2}-4 x+5=0$

(b) $x^{2}+3 x-12=0$

(c) $2 x^{2}-7 x+6=0$

(d) $3 x^{2}-6 x-2=0$

Sol :

(a) $x^{2}-4 x+5=0$

$\Rightarrow(2)^{2}-4 x^{2}+5=0$

⇒ 4 – 8 + 5 = 0

⇒ 9 – 8 ≠ 0

2 is not its root.

(b) $x^{2}+3 x-12=0$

$ \Rightarrow(2)^{2}-3 \times 2-12=0$

⇒4-6-12=4-18=-14

∴ 2 is not its roots.

(c) $2 x^{2}-7 x+6=0$

$ \Rightarrow 2(2)^{2}-7 \times 2+6=0$

⇒ 8-14+6=0 $

⇒0=0

∴ 2 is its root

(d) $3 x^{2}-6 x-2=0$

$ \Rightarrow 3(2)^{2}-6 \times 2-2=0$

⇒12-12-2=0 

⇒12-14=0

∴ 2 is not its root.

Ans : (c)

Question 4

If $\frac{1}{2}$ is a root of the equation $x^{2}+k x-\frac{5}{4}=0$ then the value of k is

(a) 2

(b) – 2

(c) $\frac{1}{4}$

(d) $\frac{1}{2}$

Sol :

$\frac{1}{2}$ is a root of the equation

$x^{2}+k x-\frac{5}{4}=0$

Substituting the value of $x=\frac{1}{2}$ in the equation 

$\left(\frac{1}{2}\right)^{2}+k \times \frac{1}{2}-\frac{5}{4}=0$

$\Rightarrow \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0$

$\Rightarrow \frac{k}{2}-1=0$

⇒k=1×2=2

∴k=2

Ans (a)

Question 5

If $\frac{1}{2}$ is a root of the quadratic equation $4 x^{2}-4 k x+k+5=0$ then the value of k is

(a) – 6

(b) – 3

(c) 3

(d) 6

Sol :

$\frac{1}{2}$ is a root of the equation 

$4 x^{2}-4 k x+k+5=0$

Substituting the value of $x=\frac{1}{2}$ in the equation

$4\left(\frac{1}{2}\right)^{2}-4 \times k \times \frac{1}{2}+k+5=0$

1-2k+k+5=0

-k+6=0

k=6

Ans (d)

Question 6

The roots of the equation $x^{2}-3 x-10=0$ are

(a) 2,- 5

(b) – 2, 5

(c) 2, 5

(d) – 2, – 5

Sol :

$x=\frac{-(-3) \pm \sqrt{(-3)^{2}-4 \times 1 \times(-10)}}{2 \times 1}$

$=\frac{3 \pm \sqrt{9+40}}{2}=\frac{3 \pm \sqrt{49}}{2}=\frac{3+7}{2}$

∴$x=\frac{3+7}{2}=5$ and $x=\frac{3-7}{2}=\frac{-4}{2}=-2$

x = 5, – 2 or – 2, 5 

Ans (b)

Question 7

If one root of a quadratic equation with rational coefficients is $\frac{3-\sqrt{5}}{2}$, then the other 

(a) $\frac{-3-\sqrt{5}}{2}$

(b) $\frac{-3+\sqrt{5}}{2}$

(c) $\frac{3+\sqrt{5}}{2}$

(d) $\frac{\sqrt{3}+5}{2}$

Sol :

One root of a quadratic equation is $\frac{3-\sqrt{5}}{2}$ then other root will be $\frac{3+\sqrt{5}}{2}$

Ans (c)

Question 8

If the equation $2 x^{2}-5 x+(k+3)=0$ has equal roots then the value of k is

(a) $\frac{g}{8}$

(b) $-\frac{g}{8}$

(c) $\frac{1}{8}$

(d) $-\frac{1}{8}$

Sol :

$2 x^{2}-5 x+(k+3)=0$

a=2, b=-5, c=k+3

=25-8(k+3)

∴ Roots are equal. 

$\therefore b^{2}-4 a c=0$

∴ 25-8(k+3)=0

⇒25-8k-24=0

⇒1-8k=0 

⇒8 k=1

$\therefore k=\frac{1}{8}$

Ans (c)

Question 9

The value(s) of k for which the quadratic equation $2 x^{2}-k x+k=0$ has equal roots is (are)

(a) 0 only

(b) 4

(c) 8 only

(d) 0, 8

Sol :

$2 x^{2}-k x+k=0$

a=2, b=-k, c=k

$\therefore b^{2}-4 a c=(-k)^{2}-4 \times 2 \times k$

$\quad=k^{2}-8 k$

∴ Roots are equal. $\therefore b^{2}-4 a c=0$

$k^{2}-8 k=0$

⇒k(k-8)=0$

Either k=0

or k-8=0, then k=8

k=0,8

Ans (d)

Question 10

If the equation $3 x^{2}-k x+2 k=0$ roots, then the the value(s) of k is (are)

(a) 6

(b) 0 Only

(c) 24 only

(d) 0

Sol :

$3 x^{2}-k x+2 k=0$

Here, a=3, b=-k, c=2 k

$b^{2}-4 a c=(-k)^{2}-4 \times 3 \times 2 k$

$=k^{2}-24 k$

∴ Roots are equal. $\therefore b^{2}-4 a c=0$

$\therefore k^{2}-24 k=0$

⇒k(k-24)=0

Either k=0 

or k-24=0, then k=24

∴ k=0, 24 

Ans (d)

Question 11

If the equation $\{k+1\} x^{2}-2(k-1) x+1=0$ has equal roots, then the values of k are

(a) 1, 3

(b) 0, 3

(c) 0, 1

(d) 0, 1

Sol :

(k + 1)x² – 2(k – 1)x + 1 = 0

Here, a = k + 1, b = -2(k – 1), c = 1

$\therefore b^{2}-4 a c=[-2(k-1)]^{2}-4(k+1)(1)$

$\quad=4\left(k^{2}-2 k+1\right)-4 k-4$

$\quad=4 k^{2}-8 k+4-4 k-4$

$\quad=4 k^{2}-12 k$

∵ Roots are equal. $\therefore b^{2}-4 a c=0$

$\therefore 4 k^{2}-12 k=0$

⇒4 k(k-3)=0 

⇒ k(k-3)=0

Either k=0 or k-3=0, then k=3

k=0,3(b)

Question 12

If the equation 2x² – 6x + p = 0 has real and different roots, then the values ofp are given by

(a) $p<\frac{9}{2}$

(b)p $\leq \frac{9}{2}$

(c) $p>\frac{9}{2}$

(d) $p \geq \frac{9}{2}$

Sol :

2x² – 6x + p = 0

Here, a = 2, b = -6, c = p

$b^{2}-4 a c=(-6)^{2}-4 \times 2 \times p$

=36-8 p

∵ Roots are real and unequal. 

$\therefore b^{2}-4 a c>0$

⇒ 36-8p>0

⇒ 36-8p>0

⇒ 36>8p

⇒ $\frac{36}{8}>p$

⇒ $p<\frac{36}{8}$

⇒ $p<\frac{9}{2}$

Ans (a)

Question 13

The quadratic equation 2x² – √5x + 1 = 0 has

(a) two distinct real roots

(b) two equal real roots

(c) no real roots

(d) more than two real roots

Sol :

2x² – √5x + 1 = 0

Here, a = 2, b = -√5, c = 1

$b^{2}-4 a c=(-\sqrt{5})^{2}-4 \times 2 \times 1$

=5-8=-3

$\because b^{2}-4 a c<0$

∵ It has no real roots.

Question 14

Which of the following equations has two distinct real roots ?

(a) $2 x^{2}-3 \sqrt{2 x}+\frac{9}{4}=0$

(b) $x^{2}+x-5=0$

(c) $x^{2}+3 x+2 \sqrt{2}=0$

(d) $5 x^{2}-3 x+1=0$

Sol :

(a) $2 x^{2}-3 \sqrt{2} x+\frac{9}{4}=0$

$b^{2}-4 a c=(-3 \sqrt{2})^{2}-4 \times 2 \times \frac{9}{4}=18-18=0$

∵ Roots are real and equal.

(b) $x^{2}+x-5=0$

$b^{2}-4 a c=(1)^{2}-4 \times 1 \times(-5)$

$=1+20=\sqrt{21}>0$

Roots are real and distinct.

Ans (b)

Question 15

Which of the following equations has no real roots ?

(a) x² – 4x + 3√2 = 0

(b) x² + 4x – 3√2 = 0

(c) x² – 4x – 3√2 = 0

(d) 3x² + 4√3x + 4 = 0

Sol :

(a) x² – 4x + 3√2 = 0

b² – 4ac = ( -4)² – 4 × 1 × 3√2

= 16 – 12√2

= 16 – 12(1.4)

= 16 – 16.8

= -0.8

b² – 4ac < 0

Roots are not real.

Ans (a)

SELINA Solution Class 9 Chapter 28 Distance Formula Exercise 28 A

Question 1.1

Find the distance between the following pairs of points:
(-3, 6) and (2, -6)

Sol:

(-3, 6) and (2, -6)
Distance between the given points
= ${\displaystyle \sqrt{{(2+3)}^2+{(-6-6)}^2}}$
= ${\displaystyle \sqrt{{\left(5\right)}^2+{(-12)}^2}}$
= ${\displaystyle \sqrt{25+144}}$
= ${\displaystyle \sqrt{169}}$
= 13

Question 1.2

Find the distance between the following pairs of points:
(-a, -b) and (a, b)

Sol:

(-a, -b) and (a, b)
Distance between the given points
= ${\displaystyle \sqrt{{(\text{a}+\text{a})}^2+{(\text{b}+\text{b})}^2}}$
= ${\displaystyle \sqrt{{\left(2\text{a}\right)}^2+{\left(2\text{b}\right)}^2}}$
= ${\displaystyle \sqrt{4{\text{a}}^2+4{\text{b}}^2}}$
= 2${\displaystyle \sqrt{{\text{a}}^2+{\text{b}}^2}}$

Question 1.3

Find the distance between the following pairs of points:
${\displaystyle (\frac{3}{5},2)\text{and}(-\frac{1}{5},1\frac{2}{5})}$

Sol:

${\displaystyle (\frac{3}{5},2)\text{and}(-\frac{1}{5},1\frac{2}{5})}$
Distance between the given points
= ${\displaystyle \sqrt{{(-\frac{1}{5}-\frac{3}{5})}^2+{(1\frac{2}{5}-2)}^2}}$
= ${\displaystyle \sqrt{{(-\frac{4}{5})}^2+{\left(\frac{7-10}{5}\right)}^2}}$
= ${\displaystyle \sqrt{\frac{16}{25}+\frac{9}{25}}}$
= ${\displaystyle \sqrt{\frac{25}{25}}}$
= 1

Question 1.4

Find the distance between the following pairs of points:
${\displaystyle (\sqrt{3}+1,1)}$ and ${\displaystyle (0,\sqrt{3})}$

Sol:

${\displaystyle (\sqrt{3}+1,1)}$ and ${\displaystyle (0,\sqrt{3})}$
Distance between the given points
= ${\displaystyle \sqrt{{(0-\sqrt{3}-1)}^2+{(\sqrt{3}-1)}^2}}$
 = ${\displaystyle \sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}}}$
= ${\displaystyle \sqrt{8}}$
= 2${\displaystyle \sqrt{2}}$

Question 2.1

Find the distance between the origin and the point:
(-8, 6) 

Sol:

Coordinates of origin are O (0, 0).
A (-8, 6)
AO = ${\displaystyle \sqrt{{(0+8)}^2+{(0-6)}^2}}$
= ${\displaystyle \sqrt{64+36}}$
= ${\displaystyle \sqrt{100}}$
= 10

Question 2.2

Find the distance between the origin and the point:
(-5, -12)

Sol:

Coordinates of origin are O (0, 0).
B (-5, -12)
BO = ${\displaystyle \sqrt{{(0+5)}^2+{(0+12)}^2}}$
= ${\displaystyle \sqrt{25+144}}$
= ${\displaystyle \sqrt{169}}$
= 13

Question 2.3

Find the distance between the origin and the point:
(8, -15)

Sol;

Coordinates of origin are O (0, 0).
C (8, -15)
CO = ${\displaystyle \sqrt{{(0-8)}^2+{(0+15)}^2}}$
= ${\displaystyle \sqrt{64+225}}$
= ${\displaystyle \sqrt{289}}$
= 17

Question 3

The distance between the points (3, 1) and (0, x) is 5. Find x.

Sol:

It is given that the distance between the points A (3, 1) and B (0, x) is 5.
∴ AB = 5
AB² = 25
(0 - 3)² + (x - 1)² = 25
9 + x² + 1 - 2x = 25
x² - 2x - 15 = 0
x² - 5x + 3x - 15 = 0
x(x - 5) + 3(x - 5)  = 0
(x - 5)(x + 3) = 0
x = 5, -3

Question 4

Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).

Sol;

Let the coordinates of the point on x-axis be (x, 0).
From the given information, we have:
${\displaystyle \sqrt{{(x-11)}^2+{(0+8)}^2}}$ = 17
(x - 11)² + (0 + 8)² = 289
x² + 121 - 22x + 64 = 289
x² - 22x - 104 = 0
x² - 26x + 4x - 104 = 0
x(x - 26) + 4(x - 26) = 0
(x - 26)(x + 4) = 0
x = 26, -4
Thus, the required co-ordinates of the points on x-axis are (26, 0) and (-4, 0).

Question 5

Find the coordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).

Sol:

Let the coordinates of the point on y-axis be (0, y).
From the given information, we have:
${\displaystyle \sqrt{{(0+8)}^2+{(y-4)}^2}}$ = 10
(0 + 8)² + (y - 4)² = 100
64 + y² + 16 - 8y = 100
y² - 8y - 20 = 0
y² - 10y + 2y - 20 = 0
y(y - 10) + 2(y - 10) = 0
(y - 10)(y + 2) = 0
y = 10, - 2
Thus, the required co-ordinates of the points on y-axis are (0, 10) and (0, -2).

Question 6

A point A is at a distance of ${\displaystyle \sqrt{10}}$ unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.

Sol:

It is given that the coordinates of point A are such that its ordinate is twice its abscissa.
So, let the coordinates of point A be (x, 2x).
We have:
${\displaystyle \sqrt{{(x-4)}^2+{(2x-3)}^2}=\sqrt{10}}$
(x - 4)² + (2x - 3)² = 10
x² + 16 - 8x + 4x² + 9 - 12x = 10
5x² - 20x + 15 = 10
x² - 4x + 3 = 0
x² - x - 3x + 3 = 0
x(x - 1) -3(x - 1) = 0
(x - 1)(x - 3) = 0
x = 1, 3
Thus, the co-ordinates of the point A are (1, 2) and (3, 6).

Question 7

A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.

Sol:

We know that the distance between the two points (x₁​, y₁​) and (x₂​, y₂​) is

d = ${\displaystyle \sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}}$

Let the given points be A = (a, 7) and B = (−3, a) and the third point given is P(2, −1).

We first find the distance between P(2, −1) and A =(a, 7) as follows:

PA = ${\displaystyle \sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}}$

​= ${\displaystyle \sqrt{{(a-2)}^2+{(7-(-1))}^2}}$

​= ${\displaystyle \sqrt{{(a-2)}^2+{(7+1)}^2}}$

​= ${\displaystyle \sqrt{{(a-2)}^2+8^2}}$

= ${\displaystyle \sqrt{{(a-2)}^2+64}}$​ 

Similarly, the distance between P(2,−1) and B = (−3, a) is:

PB = ${\displaystyle \sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}}$

= ${\displaystyle \sqrt{{(-3-2)}^2+{(a-(-1))}^2}}$

​= ${\displaystyle \sqrt{{(-5)}^2+{(a+1)}^2}}$

​= ${\displaystyle \sqrt{25+{(a+1)}^2}}$ 

Since the point P(2,−1) is equidistant from the points A(a, 7) and B = (−3, a), therefore, PA = PB that is:

${\displaystyle \sqrt{{(a-2)}^2+64}=\sqrt{25+{(a+1)}^2}}$

⇒ ${\displaystyle \left(\sqrt{{(a-2)}^2+64{}^2}\right)=(\sqrt{25+{({(a+1)}^2)}^2}}$

⇒ (a − 2)² + 64 = 25 + (a + 1)²

⇒ (a − 2)² − (a + 1)² = 25 − 64

⇒ (a² + 4 − 4a) − (a² + 1 + 2a) = −39   ...( ∵ (a − b)² = a² + b² − 2ab, (a + b)² = a² + b² + 2ab)

⇒ a² + 4 − 4a − a² − 1 − 2a = −39

⇒ −6a + 3 = −39

⇒− 6a =−39 −3

⇒ −6a = −42

⇒ a =${\displaystyle \frac{-42}{-6}}$ ​=7

Hence, a = 7.

Question 8

What point on the x-axis is equidistant from the points (7, 6) and (-3, 4)?

Sol:

Let the co-ordinates of the required point on x-axis be P (x, 0).
The given points are A (7, 6) and B (-3, 4).
Given, PA = PB
PA² = PB²
(x - 7)² + (0 - 6)² = (x + 3)² + (0 - 4)²
x² + 49 - 14x + 36 = x² + 9 + 6x + 16
60 = 20x
x = 3
Thus, the required point is (3, 0).

Question 9

Find a point on the y-axis which is equidistant from the points (5, 2) and (-4, 3).

Sol:

Let the co-ordinates of the required point on y-axis be P (0, y).
The given points are A (5, 2) and B (-4, 3).
Given, PA = PB
PA² = PB²
(0 -5)² + (y -2)² = (0 + 4)² + (y - 3)²
25 + y² + 4 - 4y = 16 + y² + 9 - 6y
2y = -4
y = -2
Thus, the required point is (0, -2).

Question 10.1

A point P lies on the x-axis and another point Q lies on the y-axis.
Write the ordinate of point P.

Sol:

Since, the point P lies on the x-axis, its ordinate is 0.

Question 10.2

A point P lies on the x-axis and another point Q lies on the y-axis.
Write the abscissa of point Q.

Sol:

Since, the point Q lies on the y-axis, its abscissa is 0.

Question 10.3

A point P lies on the x-axis and another point Q lies on the y-axis.
If the abscissa of point P is -12 and the ordinate of point Q is -16; calculate the length of line segment PQ.

Sol:

The co-ordinates of P and Q are (-12, 0) and (0, -16) respectively.
PQ = ${\displaystyle \sqrt{{(-12-0)}^2+{(0+16)}^2}}$
= ${\displaystyle \sqrt{144+256}}$
= ${\displaystyle \sqrt{400}}$
= 20

Question 11

Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.

Sol:

PQ = ${\displaystyle \sqrt{{(5-0)}^2+{(10-5)}^2}}$
= ${\displaystyle \sqrt{25+25}}$
= ${\displaystyle \sqrt{50}}$
= 5${\displaystyle \sqrt{2}}$

QR = ${\displaystyle \sqrt{{(6-5)}^2+{(3-10)}^2}}$
= ${\displaystyle \sqrt{1+49}}$
= ${\displaystyle \sqrt{50}}$
= 5${\displaystyle \sqrt{2}}$

RP = ${\displaystyle \sqrt{{(0-6)}^2+{(5-3)}^2}}$
= ${\displaystyle \sqrt{36+4}}$
= ${\displaystyle \sqrt{40}}$
= 2${\displaystyle \sqrt{10}}$

Since, PQ = QR, ΔPQR is an isosceles triangle.

Question 12

Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.

Sol :

PQ = ${\displaystyle \sqrt{{(5-0)}^2+{(10-5)}^2}}$
= ${\displaystyle \sqrt{25+25}}$
= ${\displaystyle \sqrt{50}}$
= 5${\displaystyle \sqrt{2}}$

QR = ${\displaystyle \sqrt{{(6-5)}^2+{(3-10)}^2}}$
= ${\displaystyle \sqrt{1+49}}$
= ${\displaystyle \sqrt{50}}$
= 5${\displaystyle \sqrt{2}}$

RP = ${\displaystyle \sqrt{{(0-6)}^2+{(5-3)}^2}}$
= ${\displaystyle \sqrt{36+4}}$
= ${\displaystyle \sqrt{40}}$
= 2${\displaystyle \sqrt{10}}$

Since, PQ = QR, ΔPQR is an isosceles triangle.

Sol:

PQ = ${\displaystyle \sqrt{{(6-0)}^2+{(2+4)}^2}=6\sqrt{2}\text{units}}$

QR = ${\displaystyle \sqrt{{(6-3)}^2+{(2-5)}^2}=3\sqrt{2}\text{units}}$

RS = ${\displaystyle \sqrt{{(3+3)}^2+{(5+1)}^2}=6\sqrt{2}\text{units}}$

PS = ${\displaystyle \sqrt{{(-3-0)}^2+{(-1+4)}^2}=3\sqrt{2}\text{units}}$

PR = ${\displaystyle \sqrt{{(3-0)}^2+{(5+4)}^2}=3\sqrt{10}\text{units}}$

QS = ${\displaystyle \sqrt{{(6+3)}^2+{(2+1)}^2}=3\sqrt{10}\text{units}}$

∵ PQ = RS and QR = PS,
Also PR = QS
∴ PQRS is a rectangle.

Question 13

Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.

Sol:

AB =${\displaystyle \sqrt{{(-3-1)}^2+{(0+3)}^2}=\sqrt{16+9}=\sqrt{25}}$ = 5

BC =${\displaystyle \sqrt{{(4+3)}^2+{(1+0)}^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}}$

CA =${\displaystyle \sqrt{{(1-4)}^2+{(-3-1)}^2}=\sqrt{9+16}=\sqrt{25}}$ = 5

∵ AB = CA
A, B, C are the vertices of an isosceless triangle.
AB² + CA²  = 25 + 25 = 50

BC²  = ${\displaystyle {\left(5\sqrt{2}\right)}^2}$ = 50

∴ AB² + CA² = BC²

Hence, A, B., C are the vertices of a right - angled triangle.

Hence, ΔABC is an isosceles right-angled triangle.

Area of ΔABC = ${\displaystyle \frac{1}{2}\times \text{AB}\times \text{CA}}$

= ${\displaystyle \frac{1}{2}\times 5\times 5}$

= 12.5 sq.units

Question 14

Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.

Sol:

AB = ${\displaystyle \sqrt{{(1-5)}^2+{(5-6)}^2}=\sqrt{16+1}=\sqrt{17}}$

BC = ${\displaystyle \sqrt{{(2-1)}^2+{(1-5)}^2}=\sqrt{1+16}=\sqrt{17}}$

CD = = ${\displaystyle \sqrt{{(6-2)}^2+{(2-1)}^2}=\sqrt{16+1}=\sqrt{17}}$

DA = = ${\displaystyle \sqrt{{(5-6)}^2+{(6-2)}^2}=\sqrt{1+16}=\sqrt{17}}$

AC = = ${\displaystyle \sqrt{{(2-5)}^2+{(1-6)}^2}=\sqrt{9+25}=\sqrt{34}}$

BD = = ${\displaystyle \sqrt{{(6-1)}^2+{(2-5)}^2}=\sqrt{25+9} =\sqrt{34}}$

Since, AB = BC = CD = DA and AC = BD,

A, B, C and D are the vertices of a square.

Question 15

Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

Sol:

Let the given points be A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4).

AB = ${\displaystyle \sqrt{{(-5+3)}^2+{(-5-2)}^2}=\sqrt{4+49}=\sqrt{53}}$

BC = = ${\displaystyle \sqrt{{(2+5)}^2+{(-3+5)}^2}=\sqrt{49+4}=\sqrt{53}}$

CD= = ${\displaystyle \sqrt{{(4-2)}^2+{(4+3)}^2}=\sqrt{4+49}=\sqrt{53}}$

DA = ${\displaystyle \sqrt{{(-3-4)}^2+{(2-4)}^2}=\sqrt{49+4}=\sqrt{53}}$

AC =${\displaystyle \sqrt{{(2+3)}^2+{(-3-2)}^2}=\sqrt{25+25}=5\sqrt{2}}$

BD =${\displaystyle \sqrt{{(4+5)}^2+{(4+5)}^2}=\sqrt{81+81}=9\sqrt{2}}$

Since, AB = BC = CD = DA and AC ≠ BD

The given vertices are the vertices of a rhombus.

Question 16

Points A (-3, -2), B (-6, a), C (-3, -4) and D (0, -1) are the vertices of quadrilateral ABCD; find a if 'a' is negative and AB = CD.

Sol:

AB = CD
AB² = CD²
(- 6 + 3)² + (a + 2)² = (0 + 3)² + (- 1 +  4)²
9 + a² + 4 + 4a = 9 + 9
a² + 4a - 5 = 0
a² - a + 5a - 5 = 0
a(a - 1) + 5 (a - 1) = 0
(a - 1) (a + 5) = 0
a = 1 or - 5
It is given that a is negative, thus the value of a is - 5.

Question 17

The vertices of a triangle are (5, 1), (11, 1) and (11, 9). Find the co-ordinates of the circumcentre of the triangle.

Sol:

Let the circumcentre be P (x, y).
Then, PA = PB
PA² = PB²
(x - 5)² + (y - 1)² = (x - 11)² + (y - 1)²
x² + 25 - 10x = x² + 121 - 22x
12x = 96
x = 8

Also, PA = PC
PA² = PC²
(x - 5)² + (y - 1)² = (x - 11)² + (y - 9)²
x² + 25 - 10x + y² + 1 - 2y = x² + 121 - 22x + y² + 81 - 18y
12x + 16y = 176
3x + 4y = 44
24 + 4y = 44
4y = 20
y = 5
Thus, the coordinates of the circumcentre of the triangle are (8, 5).

Question 18

Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.

Sol:

AB = 5
AB² = 25
(0 - 3)² + (y - 1 - 1)² = 25
9 + y² + 4 - 4y = 25
y² - 4y - 12 = 0
y² - 6y + 2y - 12 = 0
y(y - 6) + 2(y - 6) = 0
(y - 6) (y + 2) = 0
y = 6, -2

Question 19

Given A = (x + 2, -2) and B (11, 6). Find x if AB = 17.

Sol:

AB = 17
AB² = 289
(11 - x - 2)² + (6 + 2)² = 289
x² + 81 - 18x + 64 = 289
x² - 18x - 144 = 0
x² - 24x + 6x - 144 = 0
x(x - 24) + 6(x - 24) = 0
(x - 24) (x + 6) = 0
x = 24, -6

Question 20

The centre of a circle is (2x - 1, 3x + 1). Find x if the circle passes through (-3, -1) and the length of its diameter is 20 unit.

Sol:

Distance between the points A (2x - 1, 3x + 1) and B (- 3, - 1) = Radius of circle
AB = 10 (Since, diameter = 20 units, given)
AB² = 100
(-3 - 2x + 1)² + (-1 - 3x - 1)² = 100

(-2 - 2x)² + (-2 - 3x)² = 100

4 + 4x² + 8x + 4 + 9x² + 12x = 100

13x² + 20x - 92 = 0

x = ${\displaystyle \frac{-20\pm \sqrt{400+4784}}{26}}$

x = ${\displaystyle \frac{-20\pm 72}{26}}$

x = 2, - ${\displaystyle \frac{46}{13}}$.

Question 21

The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.

Sol:

Let the co-ordinates of point Q be (10, y).
PQ = 10
PQ² = 100
(10 - 2)² + (y + 3)² = 100
64 + y² + 9 + 6y = 100
y² + 6y - 27 = 0
y² + 9y - 3y - 27 = 0
y(y + 9) - 3(y + 9) = 0
(y + 9) (y - 3) = 0
y = -9, 3
Thus, the required co-ordinates of point Q are (10, -9) and (10, 3).

Question 22.1

Point P (2, -7) is the center of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of: AT 

Sol:

Given, radius = 13 units
PA = PB = 13 units
Using distance formula,
PT = ${\displaystyle \sqrt{{(-2-2)}^2 +{(-4+7)}^2}}$
= ${\displaystyle \sqrt{16+9}}$
= ${\displaystyle \sqrt{25}}$
= 5
Using Pythagoras theorem in Δ PAT,
AT² = PA² - PT² 
AT² = 169 - 25
AT² = 144
AT = 12 units.

Question 22.2

Point P (2, -7) is the centre of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of AB.

Sol:

We know that the perpendicular from the center of a circle to a chord bisects the chord.
∴ AB = 2AT
= 2 x 12 units
= 24 units.

Question 23

Calculate the distance between the points P (2, 2) and Q (5, 4) correct to three significant figures.

Sol:

PQ = ${\displaystyle \sqrt{{(5-2)}^2+{(4-2)}^2}}$

= ${\displaystyle \sqrt{9+4}}$

= ${\displaystyle \sqrt{13}}$

= 3.6055

= 3.61 units

Question 24

Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.

Sol:

We know that any point on x-axis has coordinates of the form (x, 0).
Abscissa of point B = 11
Since, B lies of x-axis, so its co-ordinates are (11, 0).
AB = ${\displaystyle \sqrt{{(11-7)}^2+{(0-3)}^2}}$
= ${\displaystyle \sqrt{16+9}}$
= ${\displaystyle \sqrt{25}}$
= 5 units

Question 25

Calculate the distance between A (5, -3) and B on the y-axis whose ordinate is 9.

' Sol:

We know that any point on y-axis has coordinates of the form (0, y).
Ordinate of point B = 9
Since, B lies of y-axis, so its co-ordinates are (0, 9).
AB = ${\displaystyle \sqrt{{(0-5)}^2+{(9+3)}^2}}$
= ${\displaystyle \sqrt{25+144}}$
= ${\displaystyle \sqrt{169}}$
= 13 units

Question 26

Find the point on y-axis whose distances from the points A (6, 7) and B (4, -3) are in the ratio 1: 2.

Sol:

Let the required point on y-axis be P (0, y).
PA = ${\displaystyle \sqrt{{(0-6)}^2+{(y-7)}^2}}$
= ${\displaystyle \sqrt{36+y^2+49-14y}}$
= ${\displaystyle \sqrt{y^2-14y+85}}$
PB = ${\displaystyle \sqrt{{(0-4)}^2+{(y+3)}^2}}$
= ${\displaystyle \sqrt{16+y^2+9+6y}}$
= ${\displaystyle \sqrt{y^2+6y+25}}$
From the given information, we have:
${\displaystyle \frac{\text{PA}}{\text{PB}}=\frac{1}{2}}$
${\displaystyle \frac{{\text{PA}}^2}{{\text{PB}}^2}=\frac{1}{4}}$
${\displaystyle \frac{y^2-14y+85}{y^2+6y+25}=\frac{1}{4}}$
4y² - 56y + 340
= y² + 6y + 25
3y² - 62y + 315
= 0
y = ${\displaystyle \frac{62\pm \sqrt{3844-3780}}{6}}$
y = ${\displaystyle \frac{62\pm 8}{6}}$
y = ${\displaystyle 9,\frac{35}{3}}$
Thus, the required points on y-axis are (0, 9) and ${\displaystyle (0,\frac{35}{3})}$.

Question 27

The distances of point P (x, y) from the points A (1, - 3) and B (- 2, 2) are in the ratio 2: 3.
Show that: 5x² + 5y² - 34x + 70y + 58 = 0.

Sol:

It is given that PA: PB = 2: 3

${\displaystyle \frac{\text{PA}}{\text{PB}}=\frac{2}{3}}$

${\displaystyle \frac{{\text{PA}}^2}{{\text{PB}}^2}=\frac{4}{9}}$

${\displaystyle \frac{{(x-1)}^2+{(y+3)}^2}{{(x+2)}^2+{(y-2)}^2}=\frac{4}{9}}$

${\displaystyle \frac{x^2+1-2x+y^2+9+6y}{x^2+4+4x+y^2+4-4y}=\frac{4}{9}}$

9(x² - 2x + y² + 10 + 6y) = 4(x² + 4x + y² + 8 - 4y)

9x² - 18x + 9y² + 90 + 54y = 4x² + 16x + 4y² + 32 - 16y

5x² + 5y² - 34x + 70y + 58 = 0

Hence, proved.

Question 28

The points A (3, 0), B (a, -2) and C (4, -1) are the vertices of triangle ABC right angled at vertex A. Find the value of a.

Sol:

AB = ${\displaystyle \sqrt{{(\text{a}-3)}^2+{(-2-0)}^2}}$
= ${\displaystyle \sqrt{{\text{a}}^2+9-6\text{a}+4}}$
= ${\displaystyle \sqrt{{\text{a}}^2-6\text{a}+13}}$

BC = ${\displaystyle \sqrt{{(4-\text{a})}^2+{(-1+2)}^2}}$
= ${\displaystyle \sqrt{{\text{a}}^2+16-8\text{a}+1}}$
= ${\displaystyle \sqrt{{\text{a}}^2-8\text{a}+17}}$

CA = ${\displaystyle \sqrt{{(3-4)}^2+{(0+1)}^2}}$
= ${\displaystyle \sqrt{1+1}}$
= ${\displaystyle \sqrt{2}}$

Since, triangle ABC is a right-angled at A, we have:

AB² + AC² = BC²
⇒ a² - 6a + 13 + 2 = a² - 8a + 17
⇒ 2a = 2
⇒ a = 1