write the slope intercept for the line that passes through (30,46) and is (a) parallel and (b) perpendicular to -6x-10y=7. Please show all your workNeed correct answer
We know that the equation of the line is given by:
y-y1= m(x-x1) such that m is the slope and (x1,y1) is any point of the line.
Given the point (30,46) is on the line.
==> y-46= m(x-30).
a) The line parallel to -6x-10y= 7
First we will find the slope of the parallel line by rewriting the equation in the slope-intercept form.
==> -6x -10y = 7
==> -10y = 6x +7
==> y= -(6/10)x - 7/10
Then the slope of the parallel line is -6/10 = -3/5
But we know that the slopes of two parallel lines are equal.
==> y-46 = (-3/5) (x-30)
==> y = -(3/5)x +18 + 46
==> y= -(3/5)x + 64
b) The line is perpendicular to -6x -10y = 7
We know that the slope is -3/5
Then the slope of the perpendicular line is 5/3.
==> y-46 = (5/3)(x-30)
==> y= (5/3)x -50+46
==> y= (5/3)x - 4