Find the value of x, if the points (1, x), (5, 2) and (9, 5) are colinear.
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Answers
Note down the given points
Given: (1, x), (5, 2) and (9, 5) are collinear
We know that, if the points are collinear then an area of the triangle = 0
[ \because ∵ Assume that these three points make a triangle..since
these points lies on same line, so area covered by these three point is zero.]
Step 2: Recall the area of the triangle
Area\ of\ the\ triangle=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]Area of the triangle=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
Step 3: Verify the collinear property to find unknown
Area of the triangle = 0
\frac{1}{2} [1(2 - 5) + 5(5 - x) + 9(x - 2)] = 0
2
1
[1(2−5)+5(5−x)+9(x−2)]=0
\frac{1}{2} [ -3 + 25 - 5x + 9x - 18] = 0
2
1
[−3+25−5x+9x−18]=0
\frac{1}{2}[4 +4x ] = 0
2
1
[4+4x]=0
\frac{4}{2} (1 + x) = 0
2
4
(1+x)=0
(1 + x) = 0 (1+x)=0
x = -1
Hence, the value of x = -1
Step-by-step explanation:
There are two methods by which this question can be solved:- Method 1:- For the collinear all three point must lie on the same line. Let the given 3 points be A(1,X),B(5,2),C(9,5) Consider ‘B' Let (x1=5,y1=2) Similarly for A & C (x3=1,y3=X) & (x2=9,y2=5) respectively. Lets recall the the slope point form of straight line (y-y1)=m(x-x1) Where ‘m’ is slope which is equal to (y2-y1)/(x2-x1) In this question m=5–2/9–5 =3/4. Now using slope point form: y-2=(x-5)3/4 = 4y-8=3x-15 =%3E 3x-4y-7=0 (this the equation of line) Now if we put the value of ‘x' co -oridinate then we must get the value of ‘y' co oridinate Put point A(1,X) 3–4X-7=0 =%3E -4X=4 X= -1 Method 2:- For collinear all three point must lie on same line and the slope will be equal for each and every point. i.e. (y2-y1)/(x2-x1) = (y3-y2)/(x3-x1) (5–2)/(9–5) =(X-5)/(1–9) -6=X-5 X=-1 If u get any problems comment and feel free to ask any other questions. Hope this will help you