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Zorluk: OrtaParallel and Perpendicular Lines

In the standard (x,y)(x, y) coordinate plane, square PQRSPQRS has adjacent vertices at P(1,2)P(1, 2) and Q(4,6)Q(4, 6). The line containing the side QRQR has a y-intercept of bb. What is the value of bb?

Cevap: 9

Cevap

The correct answer is 9.
The slope of segment PQPQ is 6−24−1=43\frac{6 - 2}{4 - 1} = \frac{4}{3}. Since adjacent sides of a square are perpendicular, the line containing QRQR is perpendicular to PQPQ and passes through Q(4,6)Q(4, 6). The slope of this perpendicular line is the negative reciprocal of 43\frac{4}{3}, which is −34-\frac{3}{4}. Using the slope-intercept form y=mx+by = mx + b with point Q(4,6)Q(4, 6), we substitute the values: 6=−34(4)+b⇒6=−3+b⇒b=96 = -\frac{3}{4}(4) + b \Rightarrow 6 = -3 + b \Rightarrow b = 9.

Adım Adım Çözüm

1
Calculate the slope of side PQPQ using the coordinates of P(1,2)P(1, 2) and Q(4,6)Q(4, 6).
The slope of PQPQ is mPQ=6−24−1=43m_{PQ} = \frac{6 - 2}{4 - 1} = \frac{4}{3}.
To determine the direction of side PQPQ so we can find the perpendicular slope for QRQR.
2
Find the slope of the line containing side QRQR.
The slope of QRQR is mQR=−34m_{QR} = -\frac{3}{4}.
Because adjacent sides of a square are perpendicular, the slope of QRQR is the negative reciprocal of the slope of PQPQ.
3
Find the equation of the line containing QRQR using the slope-intercept form and the coordinates of vertex Q(4,6)Q(4, 6).
Substituting the slope m=−34m = -\frac{3}{4} and point (4,6)(4, 6) into y=mx+by = mx + b gives 6=−34(4)+b6 = -\frac{3}{4}(4) + b, which simplifies to 6=−3+b6 = -3 + b, so b=9b = 9.
The line containing side QRQR must pass through vertex QQ, which allows us to determine the y-intercept bb.

Anahtar Kavram

The slopes of perpendicular lines are negative reciprocals of each other: m1⋅m2=−1m_1 \cdot m_2 = -1.
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Konular

Parallel and Perpendicular Lines

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