Advertisements
Advertisements
प्रश्न
Find the equation of the line: containing the point (2, 1) and having slope 13.
Advertisements
उत्तर
Given, slope(m) = 13 and the line passes through (2, 1).
Equation of the line in slope point form is
y – y1 = m(x – x1)
∴ the equation of the required line is
y – 1 = 13(x – 2)
∴ y – 1 = 13x – 26
∴ 13x – y = 25.
APPEARS IN
संबंधित प्रश्न
Find the equation of the line passing through the points A(2, 0) and B(3, 4).
Line y = mx + c passes through the points A(2, 1) and B(3, 2). Determine m and c.
Find the x and y-intercepts of the following line: `x/3 + y/2` = 1
Find the x and y-intercepts of the following line: 2x – 3y + 12 = 0
Find the slope, x-intercept, y-intercept of the following line : x + 2y = 0
Write the following equation in ax + by + c = 0 form: y = 2x – 4
Write the following equation in ax + by + c = 0 form: `x/2 + y/4` = 1
Find the equation of the line whose x-intercept is 3 and which is perpendicular to the line 3x – y + 23 = 0.
Reduce the equation 6x + 3y + 8 = 0 into slope-intercept form. Hence, find its slope.
Verify that A(2, 7) is not a point on the line x + 2y + 2 = 0.
Find the X-intercept of the line x + 2y – 1 = 0
Find the equation of the line: through the origin which bisects the portion of the line 3x + 2y = 2 intercepted between the co-ordinate axes.
Find the equation of the line passing through the points A(–3, 0) and B(0, 4).
The vertices of a triangle are A (1, 4), B (2, 3) and C (1, 6). Find equations of Perpendicular bisectors of sides
