Answer:
The line is;
y = 7x - 63
Explanation:
In this question, we are asked to find the equation of a line that is perpendicular to a line and passes through a specific point.
Firstly, we write the general equation of a line and define the parameters;
Generally for a line;
y = mx + c where m is the slope and c is the y-intercept
Now let’s consider the first line;
3x + 21y = 105
Let us write this in the form of the General line;
21y = -3x + 105
y = -x/7 + 5 ( I divided through by 21)
What can we identify from here? The slope or gradient of the line is -1/7
Now for a line l2 whose gradient is perpendicular to this line;
m1m2 = -1
What this means is that the product of their slopes is equal to -1
We have identified m1
So m2 will be -1/m1
This is same as m2 = -1/(-1/7) = 7
The gradient of the second line is 7
Now to write the equation of the second line, we use the following relation;
y-y1/x-x1 = m
We use the coordinates of the point through which the line passes as given in the question;
y+7/(x-8) = 7
y + 7 = 7(x-8)
y + 7 = 7x - 56
y = 7x-56-7
y = 7x - 63