WAEC - Mathematics (2019 - No. 34)
Find the equation of a straight line passing through the point (1, -5) and having gradient of \(\frac{3}{4}\)
3x + 4y - 23 = 0
3x + 4y +23 = 0
3x - 4y +23 = 0
3x - 4y - 23 = 0
Giải thích
\(\frac{3}{4} = \frac{y + 5}{x - 1}\)
3(x - 1) = 4(y + 5)
3x - 3 = 4y + 20
3x - 4y - 23 = 0