Class 10: Equation of a Line – Exercise 14(b) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the slope of the lie whose inclination is:

i) $0^o$ ii) $30^o$ iii) $72^o 30'$ iv) $46^o$

Answer:

i) $0^o$

$Slope = tan \ \theta = tan \ 0^o = 0$

ii) $30^o$

$Slope = tan \ \theta = tan \ 30^o = \frac{1}{\sqrt{3}} = 0.517$

iii) $72^o 30'$

$Slope = tan \ \theta = tan \ 72^o 30' = tan \ 72.5^o = 3.172$

iv) $46^o$

$Slope = tan \ \theta = tan \ 46^o = 1.056$

$\\$

Question 2: Find the inclination of the line whose slope is:

i) $0$ ii) $\sqrt{3}$ iii) $0.7646$ iv) $1.0875$

Answer:

i) $0$

$Slope = tan \ \theta \Rightarrow 0 = tan \ \theta \Rightarrow \theta = 0^o$

ii) $\sqrt{3}$

$Slope = tan \ \theta \Rightarrow \sqrt{3} = tan \ \theta \Rightarrow \theta = 60^o$

iii) $0.7646$

$Slope = tan \ \theta \Rightarrow 0.7646 = tan \ \theta \Rightarrow \theta = 37^o24'^o$

iv) $1.0875$

$Slope = tan \ \theta \Rightarrow 1.0875 = tan \ \theta \Rightarrow \theta = 47^o 24'$

$\\$

Question 3: Find the slope of the line passing through the following pairs of points:

i) $(-2, -3) \ and \ (1, 2)$

ii) $(-4, 0) \ and \ origin$

iii) $(a, -b) \ and \ (b, -a)$

Answer:

i) $(-2, -3) \ and \ (1, 2)$

Let $A(-2, -3) = (x_1, y_1) \ and \ B(1, 2)= (x_2, y_2)$

Therefore Slope $= \frac{y_2-y_1}{x_2-x_1}= \frac{2-(3)}{1-(-2)}= \frac{5}{3}$

ii) $(-4, 0) \ and \ origin$

Let $A(-4, 0) = (x_1, y_1) \ and \ B(0,0)= (x_2, y_2)$

Therefore Slope $= \frac{y_2-y_1}{x_2-x_1}= \frac{0-0}{0-(-4)}= 0$

iii) $(a, -b) \ and \ (b, -a)$

Let $A(a, -b) = (x_1, y_1) \ and \ B(b, -a)= (x_2, y_2)$

Therefore Slope $= \frac{y_2-y_1}{x_2-x_1}= \frac{-a-(-b)}{b-a}= 1$

$\\$

Question 4: Find the slope of the line parallel to AB if:

i) $A = (-2, 4) \ and \ B = (0, 6)$

ii) $A = (0, -3) \ and \ B = (-2, 5)$

Answer:

i) $A = (-2, 4) \ and \ B = (0, 6)$

Slope $= \frac{y_2-y_1}{x_2-x_1}= \frac{6-4}{0-(-2)}= 1$

Therefore the slope of the line parallel to $AB = 1$

ii) $A = (0, -3) \ and \ B = (-2, 5)$

Slope $= \frac{y_2-y_1}{x_2-x_1}= \frac{5-(-3)}{-2-0}= -4$

Therefore the slope of the line parallel to $AB = -4$

$\\$

Question 5: Find the slope of the line perpendicular to AB if:

i) $A = (0, -5) \ and \ B = (-2, 4)$

ii) $A = (3, -2) \ and \ B = (-1, 2)$

Answer:

i) $A = (0, -5) \ and \ B = (-2, 4)$

Slope $= \frac{y_2-y_1}{x_2-x_1}= \frac{4-(-5)}{-2-0}= \frac{-9}{2}$

Therefore the slope of the line perpendicular to $AB = \frac{-1}{\frac{-9}{2}} = \frac{2}{9}$

ii) $A = (3, -2) \ and \ B = (-1, 2)$

Slope $= \frac{y_2-y_1}{x_2-x_1}= \frac{2-(-2)}{-1-(3)}= -1$

Therefore the slope of the line parallel to $AB = \frac{-1}{-1} = 1$

$\\$

Question 6: The line passing through $(0, 2) \ and \ (-3, -1)$ is parallel to the line passing through $(-1,5) \ and \ (4, a)$ . find $a$ .

Answer:

The slope of line passing through $(0, 2) \ and \ (-3, -1) = \frac{-1-2}{-3-0} = 1$

The slope of line passing through $(-1,5) \ and \ (4, a) = \frac{a-5}{4-(-1)} = \frac{a-5}{5}$

Since the two lines are parallel to each other, their slope must be equal. Therefore

$\frac{a-5}{5} = 1 \Rightarrow a = 10$

$\\$

Question 7: The line passing through $(-4, -2) \ and \ (2, -3)$ is perpendicular to the line passing through $(a, 5) \ and \ (2, -1)$ . Find $a$ .

Answer:

The slope of line passing through $(-4, -2) \ and \ (2, -3) = \frac{-3-(-2)}{2-(-4)} = \frac{-1}{6}$

The slope of line passing through $(a, 5) \ and \ (2, -1) = \frac{-1-5}{2-(a))} = \frac{-6}{2-a}$

Since the two lines are perpendiculare to each other, the product of their slopes should be equal to $-1$ . Therefore

$\frac{-1}{6} \times \frac{-6}{2-a} = -1 \Rightarrow 6=-12+6a \Rightarrow a = 3$

$\\$

Question 8: Without using distance formula, show that the points $A (4, -2), B (-4, 4) \ and \ C (10, 6)$ are the vertices of a right-angled triangle.

Answer:

Slope of $AB = m_1 = \frac{4-(-2)}{-4-4}= \frac{-3}{4}$

Slope of $AC = m_2 = \frac{6-(-2)}{10-4}= \frac{4}{3}$

Slope of $BC = m_3 = \frac{4-6}{-4-10}= \frac{-1}{7}$

Since $m_1 \times m_2 = \frac{-3}{4} \times \frac{4}{3} = -1$

$AB$ is perpendicular to $AC$ .

Therefore $\triangle ABC$ is a right angled triangle.

$\\$

Question 9: Without using distance formula, show that the points $A (4, 5), B (1, 2), C (4, 3) \ and \ D (7, 6)$ are the vertices of a parallelogram.

Answer:

Slope of $AB = m_1 = \frac{2-5}{1-4}= \frac{-3}{3} =1$

Slope of $BC = m_2 = \frac{3-2}{4-1}= \frac{1}{3}$

Slope of $CD = m_3 = \frac{6-3}{7-4}= \frac{3}{3} = 1$

Slope of $DA = m_4 = \frac{5-6}{4-7}= \frac{-1}{-3} =\frac{1}{3}$

Therefore $m_1=m_3 \ and \ m_2=m_4$

Therefore $ABCD$ is a parallelogram.

$\\$

Question 10: $(-2, 4), (4, 8), (10, 7) \ and \ (11, -5)$ are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides is a parallelogram.

Answer:

Let $A(-2, 4), B(4, 8), C(10, 7) \ and \ D(11, -5)$ be the vertices of the quadrilateral.

Mid-point of $AB = P = (\frac{4+(-2)}{2}, \frac{8+4}{2}) = (1,6)$

Mid-point of $BC = Q = (\frac{4+10}{2}, \frac{8+7}{2}) = (7, 7.5)$

Mid-point of $DC = R = (\frac{10+11}{2}, \frac{7-5}{2}) = (10.5, 1)$

Mid-point of $AD = S = (\frac{11+(-2)}{2}, \frac{-5+4}{2}) = (4.5, -0.5)$

Slope of $PQ = m_1 = \frac{7.5-6}{7-1}= \frac{1.5}{6} =\frac{1}{4}$

Slope of $QR = m_2 = \frac{1-7.5}{10.5-7}= \frac{-6.5}{3.5}$

Slope of $RS = m_3 = \frac{-0.5-1}{4.5-10.5}= \frac{-1.5}{-6.5} = \frac{1}{4}$

Slope of $SP = m_4 = \frac{-0.5-6}{4.5-1}= \frac{-6.5}{3.5}$

Since $m_1=m_3 \ and \ m_2=m_4, \ PQRS \ is \ a \ parallelogram$

$\\$

Question 11: Show that the points $P (a, b + c), Q (b, c + a) \ and \ R (c, a + b)$ are collinear.

Answer:

$P (a, b + c), Q (b, c + a) \ and \ R (c, a + b)$

Slope of $PQ = m_1 = \frac{(c+a)-(b+c)}{b-a}= \frac{a-b}{b-a} =-1$

Slope of $RQ = m_2 = \frac{(a+b)-(c+a)}{c-b}= \frac{b-c}{c-b} = -1$

Since $m_1=m_2, P, Q, \ and \$ are collinear.

$\\$

Question 12: Find $x$ , if the slope of the line joining $(x, 2) \ and \ (8, -11)$ is $\frac{-3}{4}$ .

Answer:

Given slope of the line joining $(x, 2) \ and \ (8, -11)$ is $\frac{-3}{4}$ .

Slope of $AB \Rightarrow \frac{-3}{4} = \frac{-11-2}{8-x}$

$\Rightarrow \frac{-13}{8-x} = \frac{-3}{4}$

$\Rightarrow 52=24-3x$

$\Rightarrow x = \frac{-28}{3}$

$\\$

Question 13: The side $AB$ of an equilateral triangle $ABC$ is parallel to the $x-axis$ . Find the slopes of all the sides.

Answer:

Slope of $AB = tan \ \theta = tan \ 0^o = 0$

Slope of $BC = tan \ \theta = tan \ 120^o = -\sqrt{3} = -1.732$

Slope of $AC = tan \ \theta = tan \ 60^o = \sqrt{3} = 1.732$

$\\$

Question 14: The side $AB$ of a square $ABCD$ is parallel to the $x-axis$ . Find the slopes of all its sides. Also, find: i) The slope of the diagonal $AC$ , ii) The slope of the diagonal $BD$ .

Answer:

Slope of $AB = tan \ \theta = tan \ 0^o = 0$

Slope of $DC = slope of AB = tan \ \theta = tan \ 0^o = 0$

Slope of $BC = tan \ \theta = tan \ 90^o = \infty$

Slope of $AD = slope of BC = tan \ \theta = tan \ 0^o = 0$

Slope of $AC = tan \ \theta = tan \ 45^o = 1$

Slope of $BD = tan \ \theta = tan \ 135^o = -1$

$\\$

Question 15: $A (5, 4), B (-3, -3) \ and \ C (1, -8)$ are the vertices of a triangle $ABC$ . Find: i) The slope of the altitude of $AB$ ii) The slope of the median $AD$ and iii) The slope of the line parallel to $AC$ .

Answer:

Slope of $AB = m_1 = \frac{-2-4}{-3-5} = \frac{-6}{-8} = \frac{3}{4}$

Let slope of Altitude $= m_2$

Therefore $m_1 \times m_2 = -1$

$\Rightarrow \frac{3}{4} \times m_m=-1 \Rightarrow m_2 = \frac{-4}{3}$

Let $D$ be the midpoint of $BC$

Therefore coordinates of $D = (\frac{1-3}{2}, \frac{-8-2}{2}) = (-1, -5)$

Slope of $AD = \frac{-5-4}{1-5} = \frac{-9}{-6} = \frac{3}{2}$

Slope of $AC = \frac{-8-4}{1-5} = \frac{-12}{-4} = 3$

Therefore slope of line parallel to $AC = 3$

$\\$

Question 16: The slope of the side $BC = \frac{2}{3}$ of a rectangle $ABCD$ is . Find: i) The slope of the side $AB$ , ii) The slope of the side $AD$ .

Answer:

Since $BC \parallel AD \Rightarrow Slope \ of \ AD = \frac{2}{3}$

Since $AB \bot BC \Rightarrow Slope \ of \ AB = \frac{-1}{\frac{2}{3}} = \frac{-3}{2}$

$\\$

Question 17: Find the slope and the inclination of the line $AB$ if:

i) $A = (-3, -2) \ and \ B = (1, 2)$

ii) $A = (0, -\sqrt{3} ) \ and \ B = (3, 0)$

iii) $A = (-1, 2 \sqrt{3}) \ and \ B = (-2, \sqrt{3})$

Answer:

i) $A = (-3, -2) \ and \ B = (1, 2)$

Let $A(-3, -2) = (x_1, y_1) \ and \ B(1, 2)= (x_2, y_2)$

Therefore Slope $= \frac{y_2-y_1}{x_2-x_1}= \frac{2-(-2)}{1-(-3)}= \frac{4}{4} = 1$

Inclination : $1 = tan \ \theta \Rightarrow \theta = 45^o$

ii) $A = (0, -\sqrt{3} ) \ and \ B = (3, 0)$

Let $A(0, -\sqrt{3}) = (x_1, y_1) \ and \ B(3, 0)= (x_2, y_2)$

Therefore Slope $= \frac{y_2-y_1}{x_2-x_1}= \frac{0-(-\sqrt{3})}{3-0}= \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$

Inclination : $\frac{1}{\sqrt{3}} = tan \ \theta \Rightarrow \theta = 30^o$

iii) $A = (-1, 2 \sqrt{3}) \ and \ B = (-2, \sqrt{3})$

Let $A(-1, 2 \sqrt{3}) = (x_1, y_1) \ and \ B(-2, \sqrt{3})= (x_2, y_2)$

Therefore Slope $= \frac{y_2-y_1}{x_2-x_1}= \frac{\sqrt{3}-2\sqrt{3}}{-2-(-1)}= \sqrt{3} = 1$

Inclination : $\sqrt{3} = tan \ \theta \Rightarrow \theta = 60^o$

$\\$

Question 18: The points $(-3, 2), (2, -1) \ and \ (a, 4)$ are collinear. Find $a$ .

Answer:

$(-3, 2), (2, -1) \ and \ (a, 4)$ are collinear

$Slope \ of \ AB = Slope \ of \ BC$

$\frac{-1-2}{2-(-3)}=\frac{4-(-1)}{a-2}$

$\frac{-3}{5}=\frac{5}{a-2}$

$-3a+6=25$

$3a=-19$

$a=-\frac{19}{3}$

$\\$

Question 19: The points $(k, 3), (2, -4) \ and \ (-k+1, -2)$ are collinear. Find $K$ .

Answer:

$(k, 3), (2, -4) \ and \ (-K+1, -2)$ are collinear

$Slope \ of \ AB = Slope \ of \ BC$

$\frac{-4-3}{2-k}=\frac{-2-(-4)}{-k+1-2}$

$\frac{-7}{2-k}=\frac{2}{-k-1}$

$7k+7=4-2k$

$9k=-3$

$k=-\frac{1}{3}$

$\\$

Question 20: Plot he points $A (1, 1), B (4, 7) \ and \ C (4, 10)$ on a graph paper. Connect $A and B$ , and also $A and C$ . Which segment appears to have the steeper slope, $AB of AC$ ? Justify your conclusion by calculating the slopes of $AB \ and \ AC$ .

Answer:

Let $A(1, 1) = (x_1, y_1) \ and \ B(4, 7)= (x_2, y_2)$

Therefore Slope of $AB = \frac{y_2-y_1}{x_2-x_1}= \frac{7-1}{4-1}= 2$

Inclination : $2 = tan \ \theta \Rightarrow \theta \approx 63^o30'$

Let $A(1, 1) = (x_1, y_1) \ and \ B(4, 10)= (x_3, y_3)$

Therefore Slope of $AC = \frac{y_3-y_1}{x_3-x_1}= \frac{10-1}{4-1}= 3$

Inclination : $3 = tan \ \theta \Rightarrow \theta \approx 71^o36'$

$\\$

Question 21: Find the value(s) of $k so that$ latex PQ will be parallel to $RS. Given: i)$ latex P (2, 4), Q (3, 6), R (8, 1) \ and \ S (10, k) &s=0$

ii) $P (3, -1), Q (7, 11), R (-1, -1) \ and \ S (1, k)$

iii) $P (5, -1), Q (6, 11), R (6, -4k) \ and \ S (7, k^2)$

Answer:

i) $P (2, 4), Q (3, 6), R (8, 1) \ and \ S (10, k)$

Slope of $PQ$ = $Slope RS$

$\Rightarrow \frac{6-4}{3-2}=\frac{k-1}{10-8}$

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

$\Rightarrow k = 5$

ii) $P (3, -1), Q (7, 11), R (-1, -1) \ and \ S (1, k)$

Slope of $PQ$ = $Slope RS$

$\Rightarrow \frac{11-(-4)}{7-3}=\frac{k-(-1)}{1-(-1)}$

$\Rightarrow 3 = \frac{k+1}{2}$

$\Rightarrow k = 5$

iii) $P (5, -1), Q (6, 11), R (6, -4k) \ and \ S (7, k^2)$

Slope of $PQ$ = $Slope RS$

$\Rightarrow \frac{11-(-1)}{6-5}=\frac{k^2-(-4k)}{7-6}$

$\Rightarrow 12 = k^2+4k \Rightarrow (k+6)(k-2)$

$\Rightarrow k = -6 \ or \ 2$

$\\$

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Class 10: Equation of a Line – Exercise 14(b)

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