line through an intersection point

Suppose that the lines

$Ax+By+C=0\text{and}{A}^{\prime }x+B^{\prime }y+C^{\prime }=0$(1)

have an intersection point. Then for any real value of $k$, the equation

$Ax+By+C+k(A^{\prime }x+B^{\prime }y+C^{\prime })=0$(2)

represents a line passing through that point.

In fact, the of the equation (2) is 1, and therefore it represents a line; secondly, (2) is satisfied if both equations (1) are satisfied, and therefore the line passes through that intersection point.

Example. Determine the equation of the line passing through the point  $(-5,\mathrm{\hspace{0.17em}2})$  and the intersection point of the lines  $6x-7y+9=0$  and  $5x+9y-3=0$.

The equation of a line through the common point of those lines is

$6x-7y+9+k(5x+9y-3)=0.$(3)

We have to find such a value for $k$ that also  $(-5,\mathrm{\hspace{0.17em}2})$  lies on the line, i.e. that the equation (3) is satisfied by the values  $x=-5$,  $y=2$. So we get for determining $k$ the equation

$$-35-10k=0,$$

whence  $k=-\frac{7}{2}$. Using this value in (3), multiplying the equation by 2 and simplifying, we obtain the sought equation

$$23x+77y-39=0.$$

This result would be obtained, of course, by first calculating the intersection point of the two given lines (it is  $(-\frac{60}{89},\frac{63}{89})$) and then forming the equation of the line passing this point and the point  $(-5,\mathrm{\hspace{0.17em}2})$, but then the calculations would have been substantially longer.

Note. It is apparent that no value of $k$ allows the equation (2) to the line 
$A^{\prime }x+B^{\prime }y+C^{\prime }=0$  itself. Thus, if we had in the example instead the point  $(-5,\mathrm{\hspace{0.17em}2})$  e.g. the point  $(6,-3)$  of the line  $5x+9y-3=0$, then we had the condition  $66+0k=0$  which gives no value of $k$.