Question

show that the straight lines 4x+33y+15=0 and 33x-4y+9=0 are perpendicular ​

Answer 1

Answer: product of slope = -1

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Step-by-step explanation:

Slope of line, 4x+33y+15=0  = - coeff. of x / coeff. of y = -4/33

Slope of line, 33x-4y+9=0  =  - coeff. of x / coeff. of y = -33/-4 = 33/4

Now, product of slope is = -4/33 *  33/4 = -1

this condition shows that lines are perpendicular.

Answer 2

Answer:

LINES ARE PERPENDICULAR TO EACH OTHER.

Step-by-step explanation:

$4x + 33y + 15 = 0 \\ \\ m1(slope) = \frac{ - 33}{4} \\ \\ \\ \\ \\ 33x - 4y + 9 = 0 \\ \\ m2(slope) = - \frac{ - 4}{33} \\ \\ = \frac{4}{33} \\ \\ \\ \\ \\ \\ m1 \times m2 = - 1 \\ \\ therefore \: lines \: are \: perpendicular \: to \: each \: other.$