Straight Line and Pair of Straight Lines 1 Question 8
8. Suppose that the points $(h, k),(1,2)$ and $(-3,4)$ lie on the line $L _1$. If a line $L _2$ passing through the points $(h, k)$ and $(4,3)$ is perpendicular to $L _1$, then $k / h$ equals
(a) $-\frac{1}{7}$
(b) $\frac{1}{3}$
(c) 3
(d) 0
(2019 Main, 8 April II)
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Solution:
- Given, points $(1,2),(-3,4)$ and $(h, k)$ are lies on line $L _1$, so slope of line $L _1$ is
$$ \begin{array}{ll} & m _1=\frac{4-2}{-3-1}=\frac{k-2}{h-1} \\ \Rightarrow & m _1=\frac{-1}{2}=\frac{k-2}{h-1} \\ \Rightarrow & 2(k-2)=-1(h-1) \\ \Rightarrow & 2 k-4=-h+1 \\ \Rightarrow & h+2 k=5 \end{array} $$
and slope of line $L _2$ joining points $(h, k)$ and
$$ (4,3), \text { is } m _2=\frac{3-k}{4-h} $$
Since, lines $L _1$ and $L _2$ are perpendicular to each other.
$$ \begin{aligned} & \therefore \quad m _1 m _2=-1 \\ & \Rightarrow \quad-\frac{1}{2} \quad \frac{3-k}{4-h}=-1[\text { from Eqs. (i) and (iii)] } \\ & \Rightarrow \quad 3-k=8-2 h \\ & \Rightarrow \quad 2 h-k=5 \end{aligned} $$
On solving Eqs. (ii) and (iv), we get
$$ \begin{array}{rlrl} (h, k) & =(3,1) \\ \text { So, } & \frac{k}{h} & =\frac{3}{1}=3 \end{array} $$
