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Suppose that the lines
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(1) |
have an intersection point. Then for any real value of $k$ , the equation
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(2) |
represents a line passing through that point.
In fact, the degree 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 $(-5,\,2)$ 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
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(3) |
We have to find such a value for $k$ that also $(-5,\,2)$ 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 $(-5,\,2)$ , but then the calculations would have been substantially longer.
Note. It is apparent that no value of $k$ allows the equation (2) to represent the line
$A'x+B'y+C' = 0$ itself. Thus, if we had in the example instead the point $(-5,\,2)$ 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 $k$ .
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- K. V¨AISÄLÄ: Algebran oppi- ja esimerkkikirja II. Neljäs painos. Werner Söderström osakeyhtiö, Porvoo & Helsinki (1956).
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