Answer:
1
Step-by-step explanation:
For this question, your friend is the point-slope form.
Point-slope form = ![y-y_1 = m(x-x_1)](https://tex.z-dn.net/?f=y-y_1%20%3D%20m%28x-x_1%29)
The only variables you replace are m = slope, y1 = y of a given point, and x1 = x of a given point.
First, we need to find the slope that is perpendicular to the equation ![y=-\frac{3}{4}x+5](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B3%7D%7B4%7Dx%2B5)
An equation that is perpendicular to another equation have to be negative reciprocals (negative inverse) of each other.
If the slope is 3, then the negative reciprocal of 3 is
.
So, since the given slope is -3/4, the negative reciprocal is
.
We will use the negative reciprocal as m since we are trying to find the equation of line that includes (-3, -3) and is perpendicular to the other equation.
We will use point slope form using the (-3, -3) points and negative reciprocal as m.
![y-y_1 = m(x-x_1)\\y-(-3) = \frac{4}{3} (x-(-3))\\y+3=\frac{4}{3}(x+3)](https://tex.z-dn.net/?f=y-y_1%20%3D%20m%28x-x_1%29%5C%5Cy-%28-3%29%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%28x-%28-3%29%29%5C%5Cy%2B3%3D%5Cfrac%7B4%7D%7B3%7D%28x%2B3%29)
Now, we need to convert this into slope-intercept form. The reason why we need to do this is because y = mx + b, where b is the y-intercept. To do this, solve for y - or in other words, isolate y..
![y+3=\frac{4}{3}(x+3)\\y+3=\frac{4x}{3}+4\\y+3-3=\frac{4x}{3}+4-3\\y=\frac{4}{3}x+1](https://tex.z-dn.net/?f=y%2B3%3D%5Cfrac%7B4%7D%7B3%7D%28x%2B3%29%5C%5Cy%2B3%3D%5Cfrac%7B4x%7D%7B3%7D%2B4%5C%5Cy%2B3-3%3D%5Cfrac%7B4x%7D%7B3%7D%2B4-3%5C%5Cy%3D%5Cfrac%7B4%7D%7B3%7Dx%2B1)
The y-intercept of the line is 1.