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# (math pt slope & determinants) I have done half of this problem, is it right? and how do I do the other half?

Write the equation of a line passing through P(2,3) and Q(6,8) using

a.) the point slope formula

m= 5/4

y-3= 5/4 x - 5/2

y= 5/4x + 1/2

^^ is that the right answer??

b.) determinants

I can't figure out what to do for this one, a little help?

this is what i've done so far

x's = 2, 6

y's= 3, 8

[3(1) - 1(8)] x - [2(1) - 1(6)] y + [(2)(8) - 3(6)] 1 = 0

-5x + 4y - 2= 0

and i don't know I just get lost..

### 2 Answers

- Anonymous1 decade agoFavorite Answer
a.)

Slope-intercept form y = mx + b where m is the slope (the change in y / the change in x) and be is he y-intercept or the point where the graph of the line crosses the y axis and x = 0 and the point is (0, b).

For (2,3) and (6,8)

Yes the slope m is (8-3)/(6-2) or 5/4 which means the line will go up hill because it is a positive number

y = (5/4)x + b

SUBSTITUTE the point (2,3) into the equation and solve for b

3 = (5/4) (2) + b

3 = 5/2 + b

6/2 - 5/2 = b

1/2 = b

y = (5/4)x + 1/2 ANSWER

Your equation was correct also.

CHECK by SUBSTITUTING the other point (6,8)

8 = (5/4)(6) + 1/2

8 = 15/2 + 1/2

8 = 16/2

8 = 8 It checks out

==============================================

b.)

A line passes through the point (2, 3) and (–3, 1). Find its equation by using determinants. Set up a determinant with an x's: x, 2, and 6 in column 1, and y's: y,3, and 8 in column 2, and an all ones (1's) column 3, and set it equal to zero.

Points P(2,3) and Q(6,8)

| x y 1 |

| 2 3 1 | = 0

| 6 8 1 |

Downhill Diagonals - Uphill Diagonals = 0

(x)(3)(1) + (y)(1)(6) + (1)(2)(8) – (6)(3)(1) – (8)(1)(x) – (1)(2)(y) = 0

3x + 6y + 16 - 18 - 8x - 2y = 0

-5x +4y - 2 = 0

4y = 5x + 2

y = (5/4)x + 2/4

y = (5/4)x + 1/2 ANSWER

This is the the same equation as before

NOTE: The diagonals are written as follows

For the matrix

| a b c|

| d e f |

| g h i |

Downhill Diagonals - Uphill Diagonals =

aei + bfg + cdh - ceg - fha - ibd

GOOD LUCK

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- Anonymous1 decade ago
Slope = (8-3)/(6-2)= 5/4

y = mx + b

at point (2,3)

y = 5/4(2) + b = 3

b = 3 -10/4 = (12-10)/4 = 1/2

so, y = 5/4 x + 1/2

b) I'm not familiar with this one.

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