In what ratio is the line segment joining the points $(-1, 1)$ and $(5, 7)$ divided by the line $x + y = 4$?

The vertex $A$ of a triangle lies on the lines $x+y=1$ and $2x+3y=6$. If the orthocentre of the triangle is $O\left(\frac{3}{7}, \frac{22}{7}\right)$,then the equation of $OA$ in the normal form is

Find the equation of the line,which cuts off intercepts on the axes whose sum and product are $1$ and $-6,$ respectively.

Let $u = \hat{i} - 2\hat{j}$ and $v = -3\hat{i} + 5\hat{j}$. Consider three points $P, Q,$ and $R$ having position vectors $-\frac{1}{7}\hat{i}$,$-\frac{1}{4}\hat{j}$,and $-2\hat{i} + 3\hat{j}$ respectively. Among these,the points on the line segment passing through $u$ and $v$ are

The equation of the line whose midpoint between the axes is $({x_1}, {y_1})$ is:

Write the equation of the line passing through the points $(1, -1)$ and $(3, 5)$.