Q.14 Find the equation of the straight line passing through the point (5,7) and inclined at ${45}^0$ to x-axis. If it passes through the point P whose ordinate is -7, what is the abscissa of P?

Dear Student,

The solution to the question is as under:

$$\begin{gathered} for the equation of the line we will use slope form, i.e. \\ y=mx+c where m is the slope of line and c is the y-intercept \\ now, as slope m=\tan \theta \\ and \theta =45° \\ hence, m=\tan 45°=1 \\ So, the equation of line will be, \\ y=x+c.....\left(i\right) \\ Now, \sin ce line is pas\sin g throught (5,7), hence these coordinates must satisfy the equation of line, \\ putting the values as x=5 and y=7 in equation \left(i\right), we get, \\ 7=5+c \\ or, c=2 \\ Hence, equation of the line is, \\ y=x+2.....\left(ii\right) \end{gathered}$$
$$\begin{gathered} as the line passes through point P whose ordinate is -7 \\ and ordinate is the y coordinate of the point, \\ Hence, putting y=-7 in equation \left(ii\right), we get, \\ -7=x+2 \\ or, x=-7-2 \\ or, x=-9 \\ Thus, the abscissa of point P is -9 \end{gathered}$$

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