13. Equation of a Straight Line Previous Year Questions

Slope m = −3/4. Line through (0, 0) has c = 0 ⇒ y = (−3/4)x ⇒ 3x + 4y = 0. [1 Mark]

(c) Slope CB = (−3 − (−3)) / (0 − (−4)) = 0. Slope DA = (3 − 3) / (3 − (−1)) = 0. Since m_CB = m_DA = 0, sides CB and DA are parallel horizontal lines. [1.5 Marks]
(d) Slope of AC = (3 − (−3)) / (3 − (−4)) = 6/7. Equation: y − 3 = 6/7(x − 3) ⇒ 7y − 21 = 6x − 18 ⇒ 6x − 7y + 3 = 0. [1.5 Marks]

(a) P lies on positive x-axis ⇒ P(5, 0); Q lies on negative y-axis ⇒ Q(0, −3). [1 Mark]
(b) Slope of PQ = (−3 − 0)/(0 − 5) = 3/5.
Since RS ⊥ PQ, slope of RS = −5/3. Line RS passes through (0, 0): y = −5/3 x ⇒ 5x + 3y = 0. [2 Marks]

Correct Option: (d) 7 − 3x
Slope BC = (−5 − (−2))/(4 − 3) = −3.
Slope AB = (y + 2)/(x − 3) = −3 ⇒ y + 2 = −3x + 9 ⇒ y = 7 − 3x. [1 Mark]

(a) m₁ = k, m₂ = 3. Since perpendicular: k × 3 = −1 ⇒ k = −¹/₃. [1.5 Marks]
(b) Slope = 3, point (−1, 1): y − 1 = 3(x + 1) ⇒ y = 3x + 4. [1.5 Marks]

Part (a) Midpoint Formula: $D(x, y) = \left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}\right)$.
$D = \left(\frac{0 + (-6)}{2}, \frac{4 + 4}{2}\right) = \left(\frac{-6}{2}, \frac{8}{2}\right) = \mathbf{(-3, 4)}$.1.5 Marks

Part (b) Slope of median $AD$ passing through $A(1, -1)$ and $D(-3, 4)$:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-1)}{-3 - 1} = \frac{5}{-4} = -\frac{5}{4}$.
Equation using point-slope form $y - y_1 = m(x - x_1)$ with $A(1, -1)$:
$y - (-1) = -\frac{5}{4}(x - 1) \implies 4(y + 1) = -5(x - 1) \implies 4y + 4 = -5x + 5$.
Equation of median $AD$: $5x + 4y - 1 = 0$ (or $5x + 4y = 1$).1.5 Marks

(a) Coordinates of A and B From the graph, $A$ lies on the $x$-axis at 4 units and $B$ lies on the $y$-axis at 4 units:
$A(4, 0)$ and $B(0, 4)$.1 Mark

(b) Coordinates of P Point $P$ divides $AB$ in the ratio $m_1 : m_2 = 3 : 1$ from $A(4, 0)$ to $B(0, 4)$:
$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{3(0) + 1(4)}{3 + 1} = \frac{4}{4} = 1$
$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{3(4) + 1(0)}{3 + 1} = \frac{12}{4} = 3$
Coordinates of $P = (1, 3)$.1.5 Marks

(c) Perpendicular Line Equation Slope of line segment $AB$:
$m_{AB} = \frac{4 - 0}{0 - 4} = \frac{4}{-4} = -1$.
Since the required line is perpendicular to $AB$, its slope $m_\perp = -\frac{1}{m_{AB}} = -\frac{1}{-1} = 1$.
Equation of line with slope $m = 1$ passing through $P(1, 3)$:
$y - y_1 = m(x - x_1) \implies y - 3 = 1(x - 1) \implies y - 3 = x - 1$.
Equation of required line: $x - y + 2 = 0$ (or $y = x + 2$).1.5 Marks

Step 1 Slope of Line 1 ($5x - 3y + 2 = 0$): $m_1 = \frac{5}{3}$.
Slope of Line 2 ($6x - py + 7 = 0$): $m_2 = \frac{6}{p}$.
Since lines are perpendicular, $m_1 \times m_2 = -1$:
$$\frac{5}{3} \times \frac{6}{p} = -1 \implies \frac{10}{p} = -1 \implies \mathbf{p = -10}$$. [1.5 Marks]
Step 2 Slope of Line 2 with $p = -10$ is $m_2 = \frac{6}{-10} = -\frac{3}{5}$.
Any parallel line has the same slope $m = -\frac{3}{5}$. Passing through $(-2, -1)$:
$$y - y_1 = m(x - x_1) \implies y - (-1) = -\frac{3}{5}(x - (-2))$$
$$5(y + 1) = -3(x + 2) \implies 5y + 5 = -3x - 6 \implies \mathbf{3x + 5y + 11 = 0}$$. [1.5 Marks]

Step 1 (i) Slope of line segment $BC$ joining $B(-1, 2)$ and $C(6, -6)$:
$m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - 2}{6 - (-1)} = \frac{-8}{7} = \mathbf{-\frac{8}{7}}$. [1 Mark]
Step 2 (ii) Slope of perpendicular line ($m_\perp$):
Since $m_1 \times m_2 = -1 \implies m_\perp = -\frac{1}{m_{BC}} = -\frac{1}{-8/7} = \mathbf{\frac{7}{8}}$. [1 Mark]
Step 3 Equation of line passing through $A(3, 8)$ with slope $m = \frac{7}{8}$:
$y - y_1 = m(x - x_1) \implies y - 8 = \frac{7}{8}(x - 3)$
$8(y - 8) = 7(x - 3) \implies 8y - 64 = 7x - 21$
$\mathbf{7x - 8y + 43 = 0}$ (or $y = \frac{7}{8}x + \frac{43}{8}$). [1 Mark]

Solution For collinear points, $\text{Slope}(AB) = \text{Slope}(BC)$:
$\frac{-1 - 0}{-1 - x} = \frac{y - (-1)}{0 - (-1)} \implies \frac{1}{x + 1} = y + 1$
$(x + 1)(y + 1) = 1 \implies xy + x + y + 1 = 1 \implies x + y = -xy$.
Divide both sides by $xy$: $\frac{x}{xy} + \frac{y}{xy} = \frac{-xy}{xy} \implies \mathbf{\frac{1}{y} + \frac{1}{x} = -1}$.
Correct Answer: (a) -1 [1 Mark]

Step 1 (a) Convert line to slope-intercept form $y = mx + c$:
$3x - 4y + 7 = 0 \implies 4y = 3x + 7 \implies y = \frac{3}{4}x + \frac{7}{4}$.
$\mathbf{\text{Slope } m_1 = \frac{3}{4}}$. [1 Mark]
Step 2 (b) Find perpendicular slope $m_2$:
$m_1 \times m_2 = -1 \implies m_2 = -\frac{1}{m_1} = -\frac{1}{3/4} = -\frac{4}{3}$. [1 Mark]
Step 3 Form perpendicular line equation with y-intercept $c = 2$:
$y = m_2 x + c \implies y = -\frac{4}{3}x + 2$
$3y = -4x + 6 \implies \mathbf{4x + 3y - 6 = 0}$. [1 Mark]

Step 1 Find slope of line 1 ($3x - 5y = 7$):
$$5y = 3x - 7 \implies y = \frac{3}{5}x - \frac{7}{5} \implies m_1 = \frac{3}{5}$$ [1 Mark]
Step 2 Find slope of line 2 ($4x + ay + 9 = 0$):
$$ay = -4x - 9 \implies y = -\frac{4}{a}x - \frac{9}{a} \implies m_2 = -\frac{4}{a}$$ [1 Mark]
Step 3 Apply perpendicularity condition $m_1 \times m_2 = -1$:
$$\left(\frac{3}{5}\right) \times \left(-\frac{4}{a}\right) = -1 \implies -\frac{12}{5a} = -1$$ $$5a = 12 \implies \mathbf{a = \frac{12}{5} = 2.4}$$ [1 Mark]
Answer: $a = \frac{12}{5}$ (or $2.4$).

(i) Centroid Coordinates
$$x_G = \frac{-1 + 4 + 3}{3} = \frac{6}{3} = 2, \quad y_G = \frac{3 + 2 - 2}{3} = \frac{3}{3} = 1$$ $$\mathbf{G = (2, 1)}$$ [1.5 Marks]
(ii) Equation of Line Slope of $AC$ with $A(-1, 3)$ and $C(3, -2)$:
$$m = \frac{-2 - 3}{3 - (-1)} = -\frac{5}{4}$$ Equation through $G(2, 1)$ with slope $m = -\frac{5}{4}$:
$$y - 1 = -\frac{5}{4}(x - 2) \implies 4(y - 1) = -5x + 10$$ $$\mathbf{5x + 4y - 14 = 0}$$ [1.5 Marks]
Answers: (i) $G(2, 1)$; (ii) $5x + 4y - 14 = 0$.