find the equations of the following lines b. Gradient -4 and passing through point (2, 1)c. Passing through the points (2, -1) and (4, 2) d. Passing through the points (1, -3) and (6, -5)e. Passing through the point (5, -2) and parallel to x + 5y + 15 = 0f. Passing through the point (1, 6) and parallel to x - 3y - 2 = 0g. Passing through the point (-1, -5) and perpendicular to 3x + y + 2 = 0

You need these two basic solutions and facts to find the equation of a line:If you know the gradient [tex]m[/tex] and one point [tex](x_0,y_0)[/tex]:[tex]y-y_0=m(x-x_0)[/tex]If you know the gradient two points [tex](x_1,y_1),\ (x_2,y_2)[/tex]:[tex]\dfrac{x-x_2}{x_1-x_2}=\dfrac{y-y_2}{y_1-y_2}[/tex]The slope of a line is the coefficient m when you write it in the [tex]y=mx+q[/tex] formParallel lines have the same slopeThe slopes of perpendicular lines give -1 when multipliedWe can use this list to solve all the exercises:b)Use the first equation to get[tex]y-1=-4(x-2) \iff y=-4x+9[/tex]c)Use the second equation to get[tex]\dfrac{x-4}{2-4}=\dfrac{y-2}{-1-2} \iff \dfrac{x-4}{-2}=\dfrac{y-2}{-3}\iff 3(x-4)=2(y-2) \iff 3x-12=2y-4 \iff 2y = 3x-8 \iff y = \frac{3}{2}x-4[/tex]d) same as c)e) We derive the slope of the line by writing it as[tex]5y = -x-15 \iff y = -\dfrac{1}{5}x-3[/tex]So, the slope is -1/5. From here, it's the same as b)f) same as e)g) Again we find the slope as[tex]3x+y+2=0\iff y=-3x-2[/tex]so the slope is -3, and a perpendicular line has slope 1/3. From there, it's the same as b).