A line cuts the x - axis at A(7, 0) and the y - axis at B(0, -5). A variable line PQ is drawn perpendicular to AB. Cutting the x - axis at P and the y - axis at Q.
If AQ and BP intersect at R, the locus of R is

Prove that the lines given by the equations 3x + 2y + 5 = 0 and 3x + 2y − 1 = 0 are parallel. At what points do they intersect the x-axis? And the y-axis?

Q. Circle

x^{2} + y^{2} + 6 y = 0

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Draw the graph for the equation given below :The co-ordinates of the points where the line meets the X-axis and the Y-axis are respectively. 12x+23y=5

The correct option is C x-axis at (10, 0); y-axis at (0, 152)The graph of the given equation x2+2y3=5 would be as shown. Also, y=−3x+152 ∴intercept from graph are :X−axis:(10,0);Y−axis:(0,152)

Draw the graph for the equation given below :
The co-ordinates of the points where the line meets the $X$ -axis and the $Y$ -axis are respectively. $\frac{1}{2} x + \frac{2}{3} y = 5$

The correct option is C x-axis at (10, 0); y-axis at (0, $\frac{15}{2}$ )
The graph of the given equation $\frac{x}{2} + \frac{2 y}{3} = 5$ would be as shown.
Also,
$y = - 3 x + \frac{15}{2}$
$∴$ intercept from graph are :
$X - a x i s : (10, 0); Y - a x i s : (0, \frac{15}{2})$