I would use elimination
2x+3y = 7
2x-y = -5
Subtract those equations
4y = 12
Y = 3
Now plug in 3 for any y value
2x - 3 = -5
2x = -2
X = -1, y = 3
The slope-point form of line:
We have the points (-9, 7) and (6, 2). Substitute:
The slope-intercept form of line:
.
We have the slope m:
Pu the coordinates of the point (6, 2) to the equation:
<em>add 2 to both sides</em>
Answer:
∠MLP = 72° , ∠LJK = 22° , ∠JKL = 72° , ∠KLJ = 86°
Step-by-step explanation:
Here, given In ΔJLK and ΔMLP
Here, JK II ML, LM = MP
∠JLM = 22° and ∠LMP = 36°
Now, As angles opposite to equal sides are equal.
⇒ ∠MLP = ∠MPL = x°
Now, in ΔMLP
By <u>ANGLE SUM PROPERTY</u>: ∠MLP + ∠MPL + ∠LMP = 180°
⇒ x° + x° + 36° = 180°
⇒ 2 x = 180 - 36 = 144
or, x = 72°
⇒ ∠MLP = ∠MPL = 72°
Now,as JK II ML
⇒ ∠LJK = ∠JLM = 22° ( Alternate pair of angles)
Now, by the measure of straight angle:
∠MLP + ∠JLM + ∠JLK = 180° ( Straight angle)
⇒ 72° + 22° + ∠JLK = 180°
or, ∠JLK = 86°
In , in ΔJLK
By <u>ANGLE SUM PROPERTY</u>: ∠JKL + ∠JLK + ∠LJK = 180°
⇒ ∠JKL + 86° + 22° = 180°
⇒ ∠JKL = 180 - 108 = 72 , or ∠JKL = 72°
Hence, from above proof , ∠MLP = 72° , ∠LJK = 22° , ∠JKL = 72° ,
∠KLJ = 86°
