The equation of the line through the points (4,2) and (-8...
The equation of the line through the points (4,2) and (-8, -2) is 3y = px + q, where p and q are constants. Find the value of p.
1
2
3
9
Correct answer is A
Using the two - point from
\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)
\(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\)
\(\frac{y - 2}{-4} = \frac{x - 4}{-12}\)
\(\frac{-12(y -2)}{-4}\) = x - 4
3(y -2) = x -4
3y - 6 = x - 4
3y = x - 4 + 6
3y = x + 2...
By comparing the equations;
3y = px + , p = 1
If x = 3 and y = -1, evaluate 2(x\(^2\) - y\(^2\))...
What is the solution of the equation x2 - x - 1 + 0?...
\(\frac{d}{dx}\) cos(3x\(^2\) - 2x) is equal to...
If tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\) find sec\(\theta\)...
If x is a real number which of the following is more illustrated on the number line? ...
Find the values of x for which \( \frac{1}{2x^2 - 13x +15} \) is not defined,...
Simplify: \(\sqrt{108} + \sqrt{125} - \sqrt{75}\)...
IF the exterior angles of a quadilateral are yo, (2y + 5)o , (y + 15)o and (3y - 10)o, find yo...