Find the equation of the line perpendicular to 3y = 5-2x passing through the point (4,10) Full explanation please and I will give you brainliest only if your explanation makes sense!

Find the equation of the line perpendicular to 3y = 5-2x passing through the point (4,10)

First, convert the original equation to slope intercept form, y = mx + b:
3y = -2x + 5

Divide both sides by 3:
3y/3 = (-2x + 5)/3

Y = -2/3x + 5/3 ← this is the slope-intercept form of the original equation, 3y = -2x + 5.

Now that we have the slope of the original line, m1 = -2/3, we can find out what the equation of the line perpendicular to the original line.

By definition, “Perpendicular lines have negative reciprocal slopes.” What this means is that if you multiply the slope of the original equation (m1 = -2/3) by the slope of the line perpendicular to the original equation (m2), their product will equal to -1.

Since (m1 = - 2/3), then m2 = 3/2 because:
m1 × m2 = -1
-2/3 × 3/2 = -1.

Next, we will plug in the slope of the perpendicular line (m2 = 3/2) and the value of point (4,10) into the slope-intercept equation:

y = mx + b
y = 3/2x + b
10 = 3/2(4) + b
10 = 6 + b

Subtract 6 on both sides to solve for the y-intercept (b):

10 – 6 = 6 + b – 6
4 = b

The y-intercept of the perpendicular line is 4.

We can establish the equation of the perpendicular line as:
y = 3/2x + 4

Attached is a screenshot of both equations graphed on Desmos.

The graph shows intersection of lines y = -2/3x + 5/3 and y = 3/2x + 4 appear to be perpendicular (that is, they intersect at a 90° angle).

Yeah she/ he needs look how long that is man