SOLUTION: If K(-1, 2), L and N (1, -1) are vertices of angle KLN such that KLN=78.69°, KL intersects the axis at P, KL is produced. The inclination of KN is 0. The coordinates of M are (-3,
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-> SOLUTION: If K(-1, 2), L and N (1, -1) are vertices of angle KLN such that KLN=78.69°, KL intersects the axis at P, KL is produced. The inclination of KN is 0. The coordinates of M are (-3,
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Question 1169770: If K(-1, 2), L and N (1, -1) are vertices of angle KLN such that KLN=78.69°, KL intersects the axis at P, KL is produced. The inclination of KN is 0. The coordinates of M are (-3, -5). Calculate
(a) The gradient of KN
(b) The size of 0, the inclination of KN
(c) Show that the gradient of KL is equal to 1
(d) Determine the equation of the straight line KL in the form of y=mx + c Answer by Solver92311(821) (Show Source):
The assertion that is the point and is the point and the subsequent assertion that the inclination of is zero, are mutually contradictory. In order for a segment to have an inclination of zero, the -coordinates of the endpoints must be equal. Now if you meant that the inclination of is , you should have written: "The inclination of KN is theta". But I can only go on what you wrote. Repost this question spelling out things for which you cannot reproduce symbolically in plain text.
John
My calculator said it, I believe it, that settles it
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