One diagonal of a square is along the line 8x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the diagonal be $BD$ along the line $8x - 15y = 0$. The slope of $BD$ is $m_1 = \frac{8}{15}$.Let the vertex be $D(1, 2)$. The sides passing t

Vedclass · Accepted Answer

$23x - 7y - 9 = 0, 7x + 23y - 53 = 0$

Vedclass · Answer

(C) Let the diagonal be $BD$ along the line $8x - 15y = 0$. The slope of $BD$ is $m_1 = \frac{8}{15}$.Let the vertex be $D(1, 2)$. The sides passing through $D$ are $AD$ and $CD$.In a square,the diagonal makes an angle of $45^\circ$ with the sides.Let the slope of the sides be $m$. Then,$	an(45^\circ) = \left| \frac{m - \frac{8}{15}}{1 + m \cdot \frac{8}{15}} ight|$.$1 = \left| \frac{15m - 8}{15 + 8m} ight|$.This gives two cases:Case $1$: $15m - 8 = 15 + 8m$ $\Rightarrow 7m = 23$ $\Rightarrow m = \frac{23}{7}$.Case $2$: $15m - 8 = -(15 + 8m)$ $\Rightarrow 15m - 8 = -15 - 8m$ $\Rightarrow 23m = -7$ $\Rightarrow m = -\frac{7}{23}$.The equations of the sides passing through $(1, 2)$ are:$y - 2 = \frac{23}{7}(x - 1)$ $\Rightarrow 7y - 14 = 23x - 23$ $\Rightarrow 23x - 7y - 9 = 0$.$y - 2 = -\frac{7}{23}(x - 1)$ $\Rightarrow 23y - 46 = -7x + 7$ $\Rightarrow 7x + 23y - 53 = 0$.

One diagonal of a square is along the line $8x - 15y = 0$ and one of its vertices is $(1, 2)$. Then the equations of the sides of the square passing through this vertex are:

Similar Questions

If the slope of a straight line passing through $A(3, 2)$ is $3/4$,then the coordinates of the two points on the same line that are $5$ units away from $A$ are

The parallelism condition for two straight lines,one of which is specified by the equation $ax + by + c = 0$ and the other being represented parametrically by $x = \alpha t + \beta$ and $y = \gamma t + \delta$,is given by:

The points $A (1, 3)$ and $C (5, 1)$ are the opposite vertices of a rectangle. The equation of the line passing through the other two vertices and having a gradient of $2$ is:

The area of a triangle formed by a line with the coordinate axes is $49 \text{ sq. units}$. If the perpendicular drawn from the origin to this line makes an angle of $45^\circ$ with the positive $X$-axis, then the equation of the line is:

If the points $A(5, k)$,$B(-3, 1)$ and $C(-7, -2)$ are collinear,then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module