The points (4, 1) and (x, -6) lie on the same line. If the slope of the line is 1, what is the value of x?

Mathematics

2 answers:

svp [43]2 years ago

8 0

Answer:

Step-by-step explanation:

Calculate the slope m using the slope formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (4, 1) and (x₂, y₂ ) = (x, - 6)

m = $\frac{-6-1}{x-4}$ = 1

m = $\frac{-7}{x-4}$ = 1 ( cross- multiply )

x - 4 = - 7 ( add 4 to both sides )

x = - 3 → A

myrzilka [38]2 years ago

4 0

Slope formula is m= (y2-y1)/(x2-x1)

We know that slope is 1. So,

(-6-1)/(x-4) = 1

Then solve for x

-6-1 = x-4

x-4 = -7

x= -7+4

x= -3

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astra-53 [7]

$x+y=k\implies y=k-x$

Substitute this into the parabolic equation,

$k-x=6-(x-3)^2\implies x^2-7x+(k+3)=0$

We're told the line $x+y=k$ intersects $C$ twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know

$(-7)^2-4(k+3)=49-4(k+3)>0\implies k$

We can tell from the quadratic equation that $C$ has its vertex at the point (3, 6). Also, note that

$-1\le\sin t\le1\implies3\le3+2\sin t\le5\implies3\le x\le5$

and

$-1\le\cos2t\le1\implies2\le4+2\cos2t\le6\implies2\le y\le6$

so the furthest to the right that $C$ extends is the point (5, 2). The line $x+y=k$ passes through this point for $2+5=k\implies k=7$ . For any value of $k$ , the line $x+y=k$ passes through $C$ either only once, or not at all.