This is not possible, since the point (4,6) lies inside the circle : X2 + Y2 = 16 Tangents to a circle or ellipse never pass through the circle
If you mean: y = -23x+3 then the parallel equation is y = -23x+164
y = 1/3x+4/3
You have to draw the x and y axes, probably from -5 to 5 on each Then plot the point (-4,-2) and the point (0,2) because the intercept is two. Then draw a line between the two points, and continue it either side!
It intercepts the y axis at (0, 5) and it intercepts the x axis at (-2.3, 0) passing through the I, II and III quadrants
If a line has equation y = mx + c, the perpendicular line has gradient -1/m A line perpendicular to 3x + y = 2 has equation 3y = x + c; the value for c will be determined by a point through which the line must pass.
The equation in point slope of the line which passes through -2 -3 and is parallel to 3x plus 2y 10 is y=-1.5x.
7x-y-28 = 0
A straight line, passing through the point (0,5) with a gradient of -3.
3x-4y-6 = 0
Yes its on the line.
The graph is a straight line, with a slope of 40, passing through the point y=50 on the y-axis.
This is not possible, since the point (4,6) lies inside the circle : X2 + Y2 = 16 Tangents to a circle or ellipse never pass through the circle
There is only one line in the x-y plane and that is a straight line with a gradient of -1, passing through the point (0, 10)
If you mean: y = -23x+3 then the parallel equation is y = -23x+164
y-3 = -1/2(x-0) y = -1/2x+3
y = 1/3x+4/3