Please help! I can only fine the equation of the straight line..?

Question

A straight line through the point (0,4.5) intersects the curve x^2+xy = 45 at (6,1.5). Calculate the coordinates of the point at which the line meets the curve again.

Answer 1

Step 1

find the slope of the line

since the line passes through the point (0,4.5) and it intersects the curve at (6,1.5) we can use it to find the slope

(y2-y1) /(x2-x1) = m

(1.5 - 4.5)/(6 - 0) = m

-3/6 = m

-1/2=m ( slope is negative because it is going from left to right downward)

two points were given, but this is not a two point form because the other point is on the y - axis (0,4.5), this line is called the slope intercept form.

slope intercept from = The intercepts of a line are the directed distances from the origin to points where the line crosses the coordinate axes.

y = -(1/2)x + 4.5 equation 1

x^2+xy = 45 equation 2

substitute y of equation 1 to the y of equation 2

x^2 + x [-(1/2)x + 4.5] = 45

x^2 -(1/2) x^2 + 4.5 x = 45

[(1/2)x^2 + 4.5 x = 45] * 2 ( multiply both sides of the equation to get rid of 1/2)

x^2 + 9x = 90

using completing square

x^2 + 9x + 20.25 -20.25 = 90

x^2 + 9x + 20.25 = 90 + 20.25

(x + 4.5)^2 = 110.25

get the square root of both sides

x + 4.5 = +10.5

x + 4.5 = -10.5

at +10.5

x=10.5 - 4.5 = 6

therefore using equation 1 y = -(1/2)x + 4.5 = 1.5

(6,1.5) this is already given

at -10.5

x= -10.5 - 4.5 = - 15

using equation 1 again y = -(1/2)x + 4.5 = 12

( -15, 12) the other point of intersection

Answer 2

equation of a straight line is y =mx+c
the straight line passes through (0,4.5) and (6,1.5)
substituting for y and x using the point (0,4.5)
4.5 = m*0 +c
so c = 4.5
substitute c=4.5 and use the point (6,1.5)
so 1.5 = 6m +4.5
subtracting 4.5 from both the sides
so 6m = 1.5-4.5 = -3
so m = -3/6 = -0.5
so equation of straight line is y= -0.5x +4.5
where this line meets x^2+xy = 45?
letr that point be (xi,yi)
substituting in both equations
yi= -0.5xi+4.5
and xi^2 +xiyi =45
substituting for yi
so xi^2 + xi(-0.5xi+4.5) = 45
xi^2-0.5xi^2+4.5xi = 45
0.5xi^2+4.5xi = 45
multiplying both sides by 2
xi^2+9xi = 90
subtracting both sides by 90
xi^2+9xi-90= 0
the 2 integers with product 90 and sum +9
are +15 and -6
xi^2+9xi-90 =0
xi^2+15xi-6xi-90=0
xi(xi+15)-6(xi+15)=0
(xi-6)(xi+15)=0
xi-6= 0 or xi+15 =0
so xi=6 or xi=-15
as we have already accounted for x=6 the other x=-15 is to be taken
y = -0.5x+4.5
substituting x=15 in the above equation
y = -0.5*-15 + 4.5
= +7.5+4.5 = 12 so the coordinates of the point where the straightline again intersects the curve is (-15, 12)

Answer 3

You know two points on the line, namely, (0, 9/2) and (5, 3/2)
we may now find the equation of the line,

(y - (3/2)) / (9/2 - 3/2) = (x - 5) /(-5)
then
y - 3/2 = 15(5-x)
y = -15x + 75 - 3/2 = -15x + 147/2

Now substitute this y value into x^2 + xy = 45
the resulting equation is quadratic and may give multiple real values of x for which there will be unique values of y

Answer 4

Eqn of straight line : Y = MX + C
substituting : 4.5 = M x 0 + C
C = 4.5

therefore at 1st point; substituing X and Y values
1.5 = M x 6 + 4.5
6 x M = 1.5 - 4.5
M = -0.5

Therefore Eqn of Straight line is : Y = -0.5X + 4.5 (eqn 1)

At 2nd point subtituting eqn 1 into the curve,

X^2 + X(-0.5X + 4.5) = 45
X^2 -0.5X^2 + 4.5X -45 = 0
0.5X^2 + 4.5X -45 = 0 (MULTIPLYING BY 2 WE GET)
X^2 + 9X - 90 = 0
(X + 15 )(X - 6 )=0

THERE FORE THE 2ND POINT'S X VALUE IS -15

Substtuting into eqn 1, y will have a value of 12

ANS (-15, 12)

Answer 5

You need to find the equation of the straight line
y = mx + c
which passes through both (0, 4.5) and (6, 1.5).

Once you've done this you can plug this formula for y into x^2 + xy = 45, to get an equation which you solve for x (one solution you already have, namely x = 6 - you need the other one).

Answer 6

the final form of the equation of a quickly line is ax + by employing + c = 0 the line passes by using (3,-a million) and (-2,2), hence 3a - b + c = 0 -2a +2b + c = 0 removing c: 5a - 3b = 0 b = (5/3)a enable a equivalent 3, then we get b = 5, and c = 2a - 2b = -4 The equation of the line passing by using (3,-a million) and (-2,2) is 3x + 5y - 4 = 0 enable (x1,0) be the coordinates of the intersection of the line and x-axis, we get 3*x1 + 5*0 - 4 = 0 x1 = 4/3

Answer 7

(-15,12)

line y=-0.5x+4.5
and you replace y by the line equation in the curve
x^2+x(-0.5x+4.5)=45
0.5x^2+4.5x-45=0
2 solutions 6 and -15 (6 being for the first intersection)
so the second intersection is (-15,12)