Find the equation perpendicular to the line y=−3x+5 that passes through point (2, 6).

Given equation of line y=−3x+5.

On comparing with slope-intercept form y=mx+b, we can see that m=-3.

So the slope of the given line is -3.

We are given that lines are perpendicular.

Slopes of perpendicular line are reciprocal and opposite in sign.

We got slope of given line -3. The reciprocal of -3 is -1/3 and opposite sign of -1/3 would be 1/3.

So, the slope of the required equation of line would be 1/3.

We are given a point (2, 6).

We got a point and slope. So we can apply point-slope form directly.

y-y1 = m(x-x1)

y-6 = 1/3 (x-2).

Distributing 1/3 on right side.

y-6 = 1/3 x - 3/2

Adding 6 on both sides, we get

y-6+6 = 1/3 x - 3/2 +6

Simplifying fractions -3/2+6 by taking common denominator, we get

-3/2+6/1 = -3/2 + 12/2 = (-3+12)/2 = 9/2

Therefore, the required equation will be

y = 1/3 x +9/2