NIMCET Straight Line Previous Year Questions (PYQs) – Page 2 of 3

Given: Points: \( P(1, 4) \), \( Q(k, 3) \)

Step 1: Find midpoint of PQ

Midpoint = \( \left( \dfrac{1 + k}{2}, \dfrac{4 + 3}{2} \right) = \left( \dfrac{1 + k}{2}, \dfrac{7}{2} \right) \)

Step 2: Find slope of PQ

Slope of PQ = \( \dfrac{3 - 4}{k - 1} = \dfrac{-1}{k - 1} \)

Step 3: Slope of perpendicular bisector = negative reciprocal = \( k - 1 \)

Step 4: Use point-slope form for perpendicular bisector:

\( y - \dfrac{7}{2} = (k - 1)\left(x - \dfrac{1 + k}{2}\right) \)

Step 5: Find y-intercept (put \( x = 0 \))

\( y = \dfrac{7}{2} + (k - 1)\left( -\dfrac{1 + k}{2} \right) \)

\( y = \dfrac{7}{2} - (k - 1)\left( \dfrac{1 + k}{2} \right) \)

Given: y-intercept = -4, so:

\( \dfrac{7}{2} - \dfrac{(k - 1)(k + 1)}{2} = -4 \)

Multiply both sides by 2:

\( 7 - (k^2 - 1) = -8 \Rightarrow 7 - k^2 + 1 = -8 \Rightarrow 8 - k^2 = -8 \)

\( \Rightarrow k^2 = 16 \Rightarrow k = \pm4 \)

✅ Final Answer: $\boxed{k = -4 \text{ or } 4}$

A point equidistant from the coordinate axes satisfies:

$|x|=|y|$

So, either

$y=x$

or

$y=-x$

Given line is:

$8x-3y=11$

Case 1:

$y=x$

Put $y=x$ in the line:

$8x-3x=11$

$5x=11$

$x=\frac{11}{5}$

This is not an integer, so this case is rejected.

Case 2:

$y=-x$

Put $y=-x$ in the line:

$8x-3(-x)=11$

$8x+3x=11$

$11x=11$

$x=1$

Then,

$y=-1$

So, the point is:

$(1,-1)$

Here $x>0$ and $y<0$, so the point lies in the IV quadrant.