Question

#Quality Question

@Coordinate Geometry

■]] Suppose that the points (h,k) , (1,2) and (-3,4) lie on a line L1. If a line L2 1 passing through the points (h,k) and (4,3) is perpendicular to L1,
Then k/h equals ​

Answer 1

$\large\qquad \qquad \underline{ \pmb {{ \mathbb{ \maltese \: ANSWER :-) \: \: \text3 }}}}$

$\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: SOLUTION}}}}}$

Given points (1,2) , (3,4) and (h,k) are on Line L1,

$\underline\mathcal{ \bigstar So, \: Slope \: \: of \: \: Line \: \: L_1 \: \: is } \\ \\ m_1= \dfrac{4-2}{-3-1} = \sf { { }{} \frac{k - 2}{h - 1} } \\ \\ \implies \bf m_1 = \bf \frac{ - 1}{2 } = \bf\frac{ h- 2}{k - 1} \\ \\ \implies2(k - 2) = - 1(h - 1) \\ \\ \implies \boxed{ \boxed{ \bf h + 2k = 5}}...(i)$

And

$\underline\mathcal{ \bigstar \: Slope \: \: of \: \: Line \: \: L_2 \: \:joining } \\ \underline\mathcal{ points \: \: (h , k) \: and \: \: (4,3) \: \: is} \\ \\ \sf m_2 = \frac{3 - k}{4 - h}$

$\huge \mathcal{As}$

$\underline\mathcal{ \maltese \: Line \: \: L_1 \: \: and \: \: L_2 \: \: are \: \: } \\ \\ \underline\mathcal{ perpendicular \: \: to \: \: each \: \: other}$

$\therefore \: m_1m_2 = - 1$

$( - \dfrac{1}{2} ) ( \dfrac{3 - k}{4 - h} ) = - 1 \\ \\ \implies \boxed{\boxed{\bf2h - k = 5}} ...(ii)$

Solving (i) and (ii) we get

(h,k) = (3,1)

$\bf \dfrac{k}{h} = \dfrac{1}{3}$

Answer 2

Given :-

Suppose that the points (h,k) , (1,2) and (-3,4) lie on a line L1. If a line L2 1 passing through the points (h,k) and (4,3) is perpendicular to L1,

To Find :-

k/h

Solution :-

We know that

$\bf Slope = \dfrac{y_2-y_1}{x_2-x_1}$

$\sf m_1 = \dfrac{4-2}{-3-1} = \dfrac{k-2}{h-1}$

$\sf m_1 = \dfrac{2}{-4}=\dfrac{k-2}{h-1}$

$\sf m_1=\dfrac{-1}{2}=\dfrac{k-2}{h-1}$

Now

$\sf 2(k-2)= -1(h-1)$

$\sf 2k-4 = -h+ 1$

$\sf h +2k=5$

Now

$\sf m_1m_2=-1$

$\sf\dfrac{-1}{2}\times \dfrac{3-k}{4-h}=1$

$\sf 2(3-k) = -1(4-k)=-1$

$\sf 2h-k=5$

Now

k/h = 1/3