provide solution plz

Slope of line passing through (-6,2) and (-14, -4) is = (-4-2) / (-14+6) = 3/4
Slope of line passing through (-12,1) and (-3, -11) is = (-11-1) / (-3+12) = -4/3

Eq. of a line in XY plane:
y = mx + b [m is the slope and (x,y) is a point on that line]

Y interceptor of line passing through (-6,2) and (-14, -4) is:

2 = (3 / 4) * -6 + b
b = 26 / 4

Y interceptor of line passing through (-12,1) and (-3, -11) is:
1 = (-4 / 3) * -12 + b
b = - 45 / 3

What is the unit length of the longest side of a triangle formed by the y-axis and these two lines is = 26 / 4 + 45 / 3 = 21.5

Equation of line passing through (-6, 2) and (-14, -4);

$y = \frac{(-4-2)}{(-14+6)}x + c$

$y = \frac{3}{4}x + c$

y-intercept (c) = $y - \frac{3}{4}x$

$c = 2 - \frac{3}{4}(-6) = \frac{13}{2}$

Equation of line passing through (-12, 1) and (-3, -11);

$y = \frac{(-11-1)}{(-3+12)}x + c$

$y = \frac{-4}{3}x + c$

y-intercept (c) = $y - \frac{-4}{3}x$

$c = 1 - \frac{-4}{3}(-12) = -15$

Now, the distance between the y-intercepts = $|\frac{13}{2}| + |-15| = 21.5$

Hence, option D

Carcass

is it possible to do this with 1-1.5 minutes? or is there any trick to it?
took some time to find 2 equation and y-int

It can be observed that the longest side of the triangle is the one positioned along the Y-axis.

Applying the two-point formula to the points (-6,2) and (-14,-4), the equation of the line can be expressed as 4y - 3x - 26 = 0.

In a similar manner, for the other pair, the equation becomes 3y + 4x + 45 = 0.

To determine where each of these intersects at the Y-axis, substitute x = 0 into both equations. This results in y = -15 and y = 6.5. Therefore, the distance between these two points, which represents the longest side, is 15 + 6.5 = 21.5.

My personal opinion is one can not do more than that

Neither Bunuel on gmatclub https://gmatclub.com/forum/on-the-xy-pl ... 70981.html gave a quick solution.