Find the equation of the parallel to the line 3x-2y=5 and passing through the midpoint of the line segment joining the points (-4,2) and (2,4).​

Given:

The equation of parallel line is:

The required line passing through the midpoint of the line segment joining the points (-4,2) and (2,4).​

To find:

The equation of required line.

Solution:

Midpoint of  line segment joining the points (-4,2) and (2,4) is:

It means the required line passes through the point (-1,3).

The slope of line is:

The given equation is:

Here, A=3 and B=-2. So, the slope of the line is:

Slope of parallel lines are same. So, the slope of the required line is 1.5.

The required line passes through the point (-1,3) with slope 1.5. So, the equation of the line is:

Adding 3 on both sides, we get

Therefore, the equation of the required line is .

1. 180°. the sum of the interior angles of a triangle is 180°, for the equilateral triangle that it has three equals side is 60°+60°+60°=180°.

2. 60°. each interior angle of a equilateral triangle is 60°

3. 120°. the measure of ∠m is 120°, the sum of the angles ∠m and ∠n is equal to 180°. so, in order to find the measure of ∠m, we have ∠m+∠n=180° > ∠m=180°-∠n > ∠m=180°-60°=120°.

4. 360°. the sum of the exterior angles of an equilateral triangle is ∠m+∠r+∠x=120°+120°+120°=360°. the exterior angles of any triangle always add 360°.