What is the equation of the line perpendicular to the line 2x + 3y = -6 and having Y-intercept 3?
3x - 2y = -6
Step 1 — Slope of given line: Rewriting \(2x+3y=-6\) as \(y=-\tfrac{2}{3}x-2\) gives slope \(m_1=-\tfrac{2}{3}\).
Step 2 — Perpendicular slope: \(m_2=-\dfrac{1}{m_1}=\dfrac{3}{2}\).
Step 3 — Apply y-intercept 3:
\[y=\tfrac{3}{2}x+3 \implies 2y=3x+6 \implies 3x-2y=-6\]
Therefore the equation is 3x - 2y = -6.
The graphs of the linear equations 4x - 2y = 10 and 4x + ky = 2 intersect at a point (a, 4). The value of k is equal to:
The area (in sq. units) of the triangle formed by the graphs of 8x + 3y = 24, 2x + 8 = y and the x-axis is:
For what value of m will the system of equations 18x - 72y + 13 = 0 and 7x - my - 17 = 0 have no solution?
What is the area (in unit squares) of the triangle enclosed by the graphs of 2x + 5y = 12, x + y = 3 and the x-axis?
The graphs of the equations 3x - 20y - 2 = 0 and 11x - 5y + 61 = 0 intersect at P(a, b). What is the value of (a 2+ b 2- ab)/(a 2- b 2+ ab)?
What is the area (in sq. units) of the triangle formed by the graphs of the equations 2x + 5y - 12 = 0, x + y = 3 and y = 0?
What is the reflection of the point (5, -3) in the line y = 3?
The graphs of the equations
\(4x + \frac{1}{3}y = \frac{8}{3}\) and \(\frac{1}{2}x + \frac{3}{4}y + \frac{5}{2} = 0\) intersect at a point P. The point P also lies on the graph of the equation:
What is the area (in unit squares) of the region enclosed by the graphs of the equations
2x – 3y + 6 = 0, 4x + y = 16 and y = 0?
What is the reflection of the point (6, -1) in the line y = 2?
The graphs of the linear equations 4x - 2y = 10 and 4x + ky = 2 intersect at a point (a, 4). The value of k is equal to:
In which ratio the point (-3, p) divides the line segment joining the points (-5, -4) and (-2, 3)?
The area (in sq. units) of the triangle formed by the graphs of 8x + 3y = 24, 2x + 8 = y and the x-axis is:
In which quadrant both abscissa and ordinate are negative?
Find the slope of the line joining the points (3, -4) and (5, 2).